Lesson 13 The Mole
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| Course: | Science 10 [5 cr] - AB Ed copy 1 |
| Book: | Lesson 13 The Mole |
| Printed by: | Guest user |
| Date: | Sunday, 7 September 2025, 6:42 PM |
Introduction
Since atoms are so small, how do you know how many atoms or molecules are in a sample?
B13.1 A sample being weighed on a scale
Balanced chemical reaction equations clearly indicate the number of atoms and molecules that are required or produced during a reaction, but how can chemists perform experiments with such minute particles? There must be a way to convert the microscopic amounts into macroscopic (measurable) amounts. This way, chemists could accurately measure exact quantities.
In this lesson, you will look at the concept of the mole and learn how to convert the number of atoms of a substance into the number of moles and into the mass of the sample.
In this lesson, you will look at the concept of the mole and learn how to convert the number of atoms of a substance into the number of moles and into the mass of the sample.
Targets
By the end of this lesson, you will be able to
-
define the term “mole” as the amount of a chemical species that contains 6.02 x 1023 particles
- calculate the molar mass of compounds and molecules
- convert between the amount in moles and the mass of a substance
- apply the law of conservation of mass to balanced chemical equations
Intro Activity
How Big Is a Mole? (Not the Animal, the Other One.) @ YouTube TED-Ed ;
This video will introduce you to the concept of the mole and analogies that will help you understand exactly how big a mole is. This will help get you in the right mindset for this lesson.
What Is a Mole?
Avogadro’s constant (the concept of the mole) is a scaling factor between microscopic amounts of chemicals and macroscopic amounts of chemicals.
B13.2 Mass of powder being weighed
A mole is a unit of measurement used in chemistry. In January 2018, IUPAC updated the definition of the mole. A mole now represents a number of particles—6.02 x 1023 particles to be precise.
1 mole = 6.02 x 1023 particles.
It is a similar concept to one dozen. If you have one dozen, you have 12 items. “Dozen” is a word that represents “counting.”
The mole is also a word that represents “counting” an exact number—just a really big number!
Atoms and molecules are incredibly small, and it would be almost impossible to carry out reactions at the molecular level. Avogadro’s constant (the concept of the mole) is a scaling factor between microscopic amounts and macroscopic amounts. For example, using the following chemical reaction,
2C(s) + O2(g) → 2CO(g)
you can interpret this reaction to mean 2 atoms of carbon react with 1 molecule of oxygen to produce 2 molecules of carbon monoxide. It would not be practical to work with such small amounts in the laboratory. Instead, the quantity is scaled up using the mole concept. This reaction can also be interpreted as 2 moles of carbon atoms react with 1 mole of oxygen molecules to produce 2 moles of carbon monoxide.
1 mole = 6.02 x 1023 particles.
It is a similar concept to one dozen. If you have one dozen, you have 12 items. “Dozen” is a word that represents “counting.”
The mole is also a word that represents “counting” an exact number—just a really big number!
Atoms and molecules are incredibly small, and it would be almost impossible to carry out reactions at the molecular level. Avogadro’s constant (the concept of the mole) is a scaling factor between microscopic amounts and macroscopic amounts. For example, using the following chemical reaction,
2C(s) + O2(g) → 2CO(g)
you can interpret this reaction to mean 2 atoms of carbon react with 1 molecule of oxygen to produce 2 molecules of carbon monoxide. It would not be practical to work with such small amounts in the laboratory. Instead, the quantity is scaled up using the mole concept. This reaction can also be interpreted as 2 moles of carbon atoms react with 1 mole of oxygen molecules to produce 2 moles of carbon monoxide.
Did You Know?
© Wikimedia Commons
B13.3 Amedeo Avogadro
B13.3 Amedeo Avogadro
Although Avogadro’s number is named after the Italian scientist Amedeo Avogadro, he is not the person who proposed the concept of the mole; it was just named in honour of him.
The concept of the mole is a method of counting and is different than the concept of mass of a sample. For example, if you were told you had 1 dozen feathers and 1 dozen elephants, you would know the following:
1 dozen feathers = 12 feathers
1 dozen elephants = 12 elephants
Would 1 dozen feathers weigh the same as 1 dozen elephants? The obvious answer is no; even though you have the same number, each individual entity has a different mass.
This is true for elements as well. One hydrogen atom would not weigh the same as one argon atom or the same as one copper atom.
B13.4 Modified Bohr diagram of hydrogen
B13.5 Modified Bohr diagram of argon
B13.6 Modified Bohr diagram of copper
This is because they are each composed of a different number of subatomic particles.
It then logically follows that one mole of each of these substances would also have different masses. The mass of one mole of an element is known as its molar mass and is located on the periodic table.
Mass of 1 mole of hydrogen = 6.02 × 1023 particles = 1.01 g. (Or, the molar mass of hydrogen is 1.01 g/mol.)
Mass of 1 mole of argon = 6.02 × 1023 particles= 39.95 g. (Or, the molar mass of argon is 39.95 g/mol.)
Mass of 1 mole of copper = 6.02 × 1023 particles= 63.55 g (Or, the molar mass of copper is 63.55 g/mol)
Read This
Please read pages 107 and 108 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on the definition of a mole. Remember,
if you have any questions or you do not understand something, ask your teacher! Practice Questions
Complete the following practice question to check your understanding of the concept you just learned. Make sure you write a complete answer to the practice
question in your notes. After you have checked your answer, make corrections to your response (where necessary) to study from.- Why is the concept of the mole needed?
The concept of the mole is needed as a way to scale between microscopic amounts and macroscopic amounts of chemical substances. - If you had 1 mole of sodium atoms, how many sodium atoms would you have?
6.02 x 1023 atoms. - If you had 1 mole of zinc atoms, how many zinc atoms would you have?
6.02 x 1023 atoms - If you had 1 mole of sodium atoms and 1 mole of zinc atoms, would the two samples have the same mass? Explain.
No, even though you have the same number of atoms, each individual atom does not have the same mass and the moles will not have the same mass either. - If you had 2 moles of sodium atoms, how many sodium atoms would you have?
2 × (6.02 x 1023 particles) = 1.20 × 1024 particles.
Calculating Molar Mass
The molar mass of elements is listed on the periodic table, but how do you determine the molar mass of compounds?
B13.7 A teacher pointing to the periodic table
It is simple to determine the molar mass of an element; it is listed on the periodic table.
For example, what is the molar mass (M) of sodium, Na?
MNa = 22.99 g/mol
For determining molar mass of an element, it is very important to recall the diatoms. Diatomic elements contain two atoms, so the molar mass is doubled.
For example, what is the molar mass (M) of sodium, Na?
MNa = 22.99 g/mol
For determining molar mass of an element, it is very important to recall the diatoms. Diatomic elements contain two atoms, so the molar mass is doubled.
For example, what is the molar mass of chlorine, Cl2?
MCl2 = 2 × 35.45 g/mol
MCl2 = 70.90 g/mol
To determine the molar mass of simple binary compounds in a 1:1 ratio, you just need to add the elemental molar masses together.
For example, what is the molar mass of sodium chloride, NaCl?
MNaCl = 22.99 g/mol + 35.45 g/mol
MNaCl = 58.44 g/mol
The following methodical approach will help you to calculate molar masses for more complex substances.
- Determine the molar mass of a substance to identify the number of atoms of each element.
- Multiply the number of atoms of each element by that atom’s molar mass.
- Add all of the values together to determine the molar mass of the compound.
Examples
Work through the following examples to gain an understanding of how to calculate molar mass. Each example has a video to go with it. To play the video, click on the play icon next to the example.Determine the molar mass for water, H2O.
Determine the molar mass for ethanol, C2H5OH.
Determine the molar mass for barium cyanide, Ba(CN)2.
Determine the molar mass for iron(III) sulfate, Fe2(SO4)3.
Read This
Please read pages 108 and 109 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on the steps for calculating molar
mass. Remember, if you have any questions or you do not understand something, ask your teacher! Practice Questions
Complete the following practice question to check your understanding of the concept you just learned. Make sure you write a complete answer to the practice
question in your notes. After you have checked your answer, make corrections to your response (where necessary) to study from.- Determine the molar mass for aluminium oxide, Al2O3.
- Determine the molar mass for calcium phosphate, Ca3(PO4)2.
Calculating Moles and Mass
If a chemical reaction requires one mole of copper atoms, does that mean you need to count out 6.02 × 1023 copper atoms?
B13.8 Jar of jelly beans
As you could imagine, it would take a long time to count out 6.02 × 1023 atoms! There must be a more efficient way. You may have seen contests where you need to guess the number of jelly beans in a jar. Can you think of how you could determine
the number of jelly beans without actually counting all of them? You may have suggested weighing the mass of one jelly bean, then weighing the mass of the whole container of jelly beans, and then—remembering to subtract the mass of the container—dividing
the total mass of all jelly beans by the mass of one jelly bean. This will give you the number of jelly beans in the container!
Chemists use this process when they calculate the amount of chemical substances. You already learned how to calculate the molar mass of a substance; now you need to be able to convert between the amount (number of moles) of a substance and its mass.
When calculating the mass in a sample, the amount (number of moles) of the sample is multiplied by the molar mass of the substance.
Chemists use this process when they calculate the amount of chemical substances. You already learned how to calculate the molar mass of a substance; now you need to be able to convert between the amount (number of moles) of a substance and its mass.
When calculating the mass in a sample, the amount (number of moles) of the sample is multiplied by the molar mass of the substance.
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Guidelines for Significant Digits, Manipulation of Data, and Rounding in Science
When completing these calculations, you will be asked to answer with the proper number of significant digits.- For all non-logarithmic values, regardless of decimal position, any of the digits 1 to 9 is a significant digit; 0 may be significant. For example:
123 0.123 0.002 30 2.30 × 103 2.03
All of these have three significant digits.
- Leading zeros are not significant. For example:
0.12 and 0.012 each have two significant digits.
- All trailing zeros are significant. For example:
200 has three significant digits.
0.123 00 and 20.000 each have five significant digits.
- When adding or subtracting measured quantities, the calculated answer should be rounded to the same degree of precision as that of the least precise number used in the computation if this is the only operation. For example in the following addition:
- When multiplying or dividing measured quantities, the calculated answer should be rounded to the same number of significant digits as are contained in the quantity with the fewest number of significant digits if this is the only operation.
For example:
(1.23)(54.321) = 66.814 83
The answer should be rounded to 66.8.
- When a series of calculations is performed, each interim value should not be rounded before carrying out the next calculation. The final answer should then be rounded to the same number of significant digits as are contained in the quantity
in the
original data with the fewest number of significant digits. For example:
In determining the value of (1.23)(4.321) / (3.45 − 3.21), three calculations are required
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The answer should be rounded to 24.8.
-
3.45 − 3.21 = 0.24
-
(1.23)(4.321) = 5.314 83
-
5.314 83 / 0.24 = 22.145 125
[Not 5.31 / 0.24 = 22.125]
The value should be rounded to 22.1.
Note: In the example given, steps a and b yield intermediate values. These values should not be used in determining the number of significant digits.
- When calculations involve exact numbers (counted and defined values), the calculated answer should be rounded based upon the precision of the measured value(s). For example:
12 eggs × 52.3 g/egg = 627.6 g
or
5 mol × 32.06 g/mol = 160.30 g
or
(1 mol)(–1 095.8 kJ/mol) + (2 mol)(40.8 kJ/mol) = –1 014.2 kJ
or
5 mol × 32.06 g/mol = 160.30 g
or
(1 mol)(–1 095.8 kJ/mol) + (2 mol)(40.8 kJ/mol) = –1 014.2 kJ
- When the first digit to be dropped is less than or equal to 4, the last digit retained should not be changed. For example:
1.234 5 rounded to three digits is 1.23.
- When the first digit to be dropped is greater than or equal to 5, the last digit retained should be increased by one. For example:
12.25 rounded to three digits is 12.3.
Examples
Each example has a video to go with it. To play the video, click on the play icon next to the example.Write the chemical formula for the substance and determine its molar mass.
Water is H2O.
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Water is H2O.
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Identify the variables from the question.
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Substitute the values into the formula.
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«math»«msub»«mi»m«/mi»«mrow»«msub»«mi»H«/mi»«mn»2«/mn»«/msub»«mi»O«/mi»«/mrow»«/msub»«mo»=«/mo»«mi»n«/mi»«mo»§#215;«/mo»«mi»M«/mi»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»m«/mi»«mrow»«msub»«mi»H«/mi»«mn»2«/mn»«/msub»«mi»O«/mi»«/mrow»«/msub»«mo»=«/mo»«mn»2«/mn»«mo».«/mo»«mn»00«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«mo»§#215;«/mo»«mn»18«/mn»«mo».«/mo»«mn»02«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»m«/mi»«mrow»«msub»«mi»H«/mi»«mn»2«/mn»«/msub»«mi»O«/mi»«/mrow»«/msub»«mo»=«/mo»«mn»36«/mn»«mo».«/mo»«mn»04«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»=«/mo»«mn»36«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#160;«/mo»«mo»(«/mo»«mi»to«/mi»«mo»§#160;«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi»sig«/mi»«mo»§#160;«/mo»«mi»digs«/mi»«mo»)«/mo»«/math»
When calculating the amount (number of moles) in a sample, the mass of the sample is divided by the molar mass of the substance.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi mathcolor=¨#FFFFFF¨»n«/mi»«/mtd»«mtd»«mo mathcolor=¨#FFFFFF¨»=«/mo»«/mtd»«mtd»«mfrac mathcolor=¨#FFFFFF¨»«mi»m«/mi»«mi»M«/mi»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mi mathcolor=¨#FFFFFF¨»n«/mi»«/mtd»«mtd»«mo mathcolor=¨#FFFFFF¨»=«/mo»«/mtd»«mtd»«mi mathcolor=¨#FFFFFF¨»amount«/mi»«mo mathcolor=¨#FFFFFF¨»§#160;«/mo»«mo mathcolor=¨#FFFFFF¨»(«/mo»«mi mathcolor=¨#FFFFFF¨»mol«/mi»«mo mathcolor=¨#FFFFFF¨»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi mathcolor=¨#FFFFFF¨»m«/mi»«/mtd»«mtd»«mo mathcolor=¨#FFFFFF¨»=«/mo»«/mtd»«mtd»«mi mathcolor=¨#FFFFFF¨»mass«/mi»«mo mathcolor=¨#FFFFFF¨»§#160;«/mo»«mo mathcolor=¨#FFFFFF¨»(«/mo»«mi mathvariant=¨normal¨ mathcolor=¨#FFFFFF¨»g«/mi»«mo mathcolor=¨#FFFFFF¨»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi mathcolor=¨#FFFFFF¨»M«/mi»«/mtd»«mtd»«mo mathcolor=¨#FFFFFF¨»=«/mo»«/mtd»«mtd»«mi mathcolor=¨#FFFFFF¨»molar«/mi»«mo mathcolor=¨#FFFFFF¨»§#160;«/mo»«mi mathcolor=¨#FFFFFF¨»mass«/mi»«mo mathcolor=¨#FFFFFF¨»§#160;«/mo»«mo mathcolor=¨#FFFFFF¨»(«/mo»«mi mathvariant=¨normal¨ mathcolor=¨#FFFFFF¨»g«/mi»«mo mathcolor=¨#FFFFFF¨»/«/mo»«mi mathcolor=¨#FFFFFF¨»mol«/mi»«mo mathcolor=¨#FFFFFF¨»)«/mo»«/mtd»«/mtr»«/mtable»«/math»
What is the amount in moles of a 5.0 g sample of carbon dioxide?
Write the chemical formula for the substance and determine its molar mass.
Carbon dioxide is CO2.
«math»«msub»«mi»M«/mi»«mrow»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mo»(«/mo»«mn»1«/mn»«mo»§#215;«/mo»«mn»12«/mn»«mo».«/mo»«mn»01«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mo»+«/mo»«mo»(«/mo»«mn»2«/mn»«mo»§#215;«/mo»«mn»16«/mn»«mo».«/mo»«mn»00«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»M«/mi»«mrow»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»44«/mn»«mo».«/mo»«mn»01«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/math»
Carbon dioxide is CO2.
«math»«msub»«mi»M«/mi»«mrow»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mo»(«/mo»«mn»1«/mn»«mo»§#215;«/mo»«mn»12«/mn»«mo».«/mo»«mn»01«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mo»+«/mo»«mo»(«/mo»«mn»2«/mn»«mo»§#215;«/mo»«mn»16«/mn»«mo».«/mo»«mn»00«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»M«/mi»«mrow»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»44«/mn»«mo».«/mo»«mn»01«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/math»
Identify the variables from the question.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»m«/mi»«mrow»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»5«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»44«/mn»«mo».«/mo»«mn»01«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»m«/mi»«mrow»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»5«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»44«/mn»«mo».«/mo»«mn»01«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
Substitute the values into the formula.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mi»m«/mi»«mi»M«/mi»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»5«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«mrow»«mn»44«/mn»«mo».«/mo»«mn»01«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»13«/mn»«mo»§#8230;«/mo»«mi»mol«/mi»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»11«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«mo»§#160;«/mo»«mo»(«/mo»«mi»to«/mi»«mo»§#160;«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mi»sig«/mi»«mo»§#160;«/mo»«mi»digs«/mi»«mo»)«/mo»«/mtd»«/mtr»«/mtable»«/math»
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mi»m«/mi»«mi»M«/mi»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»5«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«mrow»«mn»44«/mn»«mo».«/mo»«mn»01«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«msub»«mi»O«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»13«/mn»«mo»§#8230;«/mo»«mi»mol«/mi»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»11«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«mo»§#160;«/mo»«mo»(«/mo»«mi»to«/mi»«mo»§#160;«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mi»sig«/mi»«mo»§#160;«/mo»«mi»digs«/mi»«mo»)«/mo»«/mtd»«/mtr»«/mtable»«/math»
Work through the following examples to reinforce your understanding of converting between mass and moles for chemical substances.
What is the amount in moles of a 150 g sample of calcium chloride?
Write the chemical formula for the substance and determine its molar mass.
«math»«msup»«mi»Ca«/mi»«mrow»«mn»2«/mn»«mo»+«/mo»«/mrow»«/msup»«mo»+«/mo»«msup»«mi»Cl«/mi»«mo»-«/mo»«/msup»«/math»
«math» «msub» «mi»M«/mi» «mrow» «mi»C«/mi» «mi»a«/mi» «mi»C«/mi» «msub» «mi»l«/mi» «mn»2«/mn» «/msub» «/mrow» «/msub» «mo»=«/mo» «mo»(«/mo» «mn»1«/mn» «mo».«/mo» «mn»40«/mn» «mo».«/mo» «mn»08«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»g«/mi» «mo»/«/mo» «mi»mol«/mi» «mo»)«/mo» «mo»+«/mo» «mo»(«/mo» «mn»2«/mn» «mo»§#215;«/mo» «mn»35«/mn» «mo».«/mo» «mn»45«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»g«/mi» «mo»/«/mo» «mi»mol«/mi» «mo»)«/mo» «mspace linebreak=¨newline¨»«/mspace» «msub» «mi»M«/mi» «mrow» «mi»C«/mi» «mi»a«/mi» «mi»C«/mi» «msub» «mi»l«/mi» «mn»2«/mn» «/msub» «/mrow» «/msub» «mo»=«/mo» «mn»110«/mn» «mo».«/mo» «mn»98«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»g«/mi» «mo»/«/mo» «mi»mol«/mi» «/math»
«math»«msup»«mi»Ca«/mi»«mrow»«mn»2«/mn»«mo»+«/mo»«/mrow»«/msup»«mo»+«/mo»«msup»«mi»Cl«/mi»«mo»-«/mo»«/msup»«/math»
«math» «msub» «mi»M«/mi» «mrow» «mi»C«/mi» «mi»a«/mi» «mi»C«/mi» «msub» «mi»l«/mi» «mn»2«/mn» «/msub» «/mrow» «/msub» «mo»=«/mo» «mo»(«/mo» «mn»1«/mn» «mo».«/mo» «mn»40«/mn» «mo».«/mo» «mn»08«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»g«/mi» «mo»/«/mo» «mi»mol«/mi» «mo»)«/mo» «mo»+«/mo» «mo»(«/mo» «mn»2«/mn» «mo»§#215;«/mo» «mn»35«/mn» «mo».«/mo» «mn»45«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»g«/mi» «mo»/«/mo» «mi»mol«/mi» «mo»)«/mo» «mspace linebreak=¨newline¨»«/mspace» «msub» «mi»M«/mi» «mrow» «mi»C«/mi» «mi»a«/mi» «mi»C«/mi» «msub» «mi»l«/mi» «mn»2«/mn» «/msub» «/mrow» «/msub» «mo»=«/mo» «mn»110«/mn» «mo».«/mo» «mn»98«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»g«/mi» «mo»/«/mo» «mi»mol«/mi» «/math»
List the variables.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»l«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»m«/mi»«mrow»«mi»C«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»l«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»150«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«mi»C«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»l«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»110«/mn»«mo».«/mo»«mn»98«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»l«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»m«/mi»«mrow»«mi»C«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»l«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»150«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«mi»C«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»l«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»110«/mn»«mo».«/mo»«mn»98«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
Identify the correct formula and rearrange if necessary.
«math»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»l«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mfrac»«mi»m«/mi»«mi»M«/mi»«/mfrac»«/math»
«math»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»l«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mfrac»«mi»m«/mi»«mi»M«/mi»«/mfrac»«/math»
Substitute the values into the formula.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»l«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mi»m«/mi»«mi»M«/mi»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»l«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»150«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«mrow»«mn»110«/mn»«mo».«/mo»«mn»98«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»l«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mi»m«/mi»«mi»M«/mi»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»l«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»150«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«mrow»«mn»110«/mn»«mo».«/mo»«mn»98«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
Calculate the answer.
«math»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»l«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»1«/mn»«mo».«/mo»«mn»351«/mn»«mo»§#8230;«/mo»«mi»mol«/mi»«mo»=«/mo»«mn»1«/mn»«mo».«/mo»«mn»35«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«mo»§#160;«/mo»«mo»(«/mo»«mi»to«/mi»«mo»§#160;«/mo»«mi»three«/mi»«mo»§#160;«/mo»«mi»significant«/mi»«mo»§#160;«/mo»«mi»digits«/mi»«mo»)«/mo»«/math»
«math»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«mi»a«/mi»«mi»C«/mi»«msub»«mi»l«/mi»«mn»2«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»1«/mn»«mo».«/mo»«mn»351«/mn»«mo»§#8230;«/mo»«mi»mol«/mi»«mo»=«/mo»«mn»1«/mn»«mo».«/mo»«mn»35«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«mo»§#160;«/mo»«mo»(«/mo»«mi»to«/mi»«mo»§#160;«/mo»«mi»three«/mi»«mo»§#160;«/mo»«mi»significant«/mi»«mo»§#160;«/mo»«mi»digits«/mi»«mo»)«/mo»«/math»
What is the mass in grams of a 0.55 mol sample of dinitrogen trioxide?
Write the chemical formula for the substance and determine its molar mass.
«math»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»M«/mi»«mrow»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mo»(«/mo»«mn»2«/mn»«mo»§#215;«/mo»«mn»14«/mn»«mo».«/mo»«mn»01«/mn»«mo»§#160;«/mo»«mi»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mo»+«/mo»«mo»(«/mo»«mn»3«/mn»«mo»§#215;«/mo»«mn»16«/mn»«mo».«/mo»«mn»00«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»M«/mi»«mrow»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»76«/mn»«mo».«/mo»«mn»02«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/math»
«math»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»M«/mi»«mrow»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mo»(«/mo»«mn»2«/mn»«mo»§#215;«/mo»«mn»14«/mn»«mo».«/mo»«mn»01«/mn»«mo»§#160;«/mo»«mi»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mo»+«/mo»«mo»(«/mo»«mn»3«/mn»«mo»§#215;«/mo»«mn»16«/mn»«mo».«/mo»«mn»00«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»M«/mi»«mrow»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»76«/mn»«mo».«/mo»«mn»02«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/math»
List the variables.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»m«/mi»«mrow»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»55«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»76«/mn»«mo».«/mo»«mn»02«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»m«/mi»«mrow»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»55«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»76«/mn»«mo».«/mo»«mn»02«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
Identify the correct formula and rearrange if necessary.
«math»«msub»«mi»m«/mi»«mrow»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mi»n«/mi»«mo»§#215;«/mo»«mi»M«/mi»«/math»
«math»«msub»«mi»m«/mi»«mrow»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mi»n«/mi»«mo»§#215;«/mo»«mi»M«/mi»«/math»
Substitute the values into the formula.
«math»«msub»«mi»m«/mi»«mrow»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»55«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«mo»§#215;«/mo»«mn»76«/mn»«mo».«/mo»«mn»02«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/math»
«math»«msub»«mi»m«/mi»«mrow»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»55«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«mo»§#215;«/mo»«mn»76«/mn»«mo».«/mo»«mn»02«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/math»
Calculate the answer.
«math»«msub»«mi»m«/mi»«mrow»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»41«/mn»«mo».«/mo»«mn»8«/mn»«mo»§#8230;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»=«/mo»«mn»42«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#160;«/mo»«mo»(«/mo»«mi»to«/mi»«mo»§#160;«/mo»«mi»two«/mi»«mo»§#160;«/mo»«mi»significant«/mi»«mo»§#160;«/mo»«mi»digits«/mi»«mo»)«/mo»«/math»
«math»«msub»«mi»m«/mi»«mrow»«msub»«mi»N«/mi»«mn»2«/mn»«/msub»«msub»«mi»O«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»41«/mn»«mo».«/mo»«mn»8«/mn»«mo»§#8230;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»=«/mo»«mn»42«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#160;«/mo»«mo»(«/mo»«mi»to«/mi»«mo»§#160;«/mo»«mi»two«/mi»«mo»§#160;«/mo»«mi»significant«/mi»«mo»§#160;«/mo»«mi»digits«/mi»«mo»)«/mo»«/math»
What is the mass in grams of a 0.0255 mol sample of iron(III) fluoride?
Write the chemical formula for the substance and determine its molar mass.
Fe3+ F–
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«mi»F«/mi»«mi»e«/mi»«msub»«mi»F«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»(«/mo»«mn»1«/mn»«mo»§#215;«/mo»«mn»55«/mn»«mo».«/mo»«mn»85«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mo»+«/mo»«mo»(«/mo»«mn»3«/mn»«mo»§#215;«/mo»«mn»19«/mn»«mo».«/mo»«mn»00«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«mi»F«/mi»«mi»e«/mi»«msub»«mi»F«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»112«/mn»«mo».«/mo»«mn»85«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
Fe3+ F–
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«mi»F«/mi»«mi»e«/mi»«msub»«mi»F«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»(«/mo»«mn»1«/mn»«mo»§#215;«/mo»«mn»55«/mn»«mo».«/mo»«mn»85«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mo»+«/mo»«mo»(«/mo»«mn»3«/mn»«mo»§#215;«/mo»«mn»19«/mn»«mo».«/mo»«mn»00«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«mi»F«/mi»«mi»e«/mi»«msub»«mi»F«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»112«/mn»«mo».«/mo»«mn»85«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
List the variables.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»m«/mi»«mrow»«mi»F«/mi»«mi»e«/mi»«msub»«mi»F«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»F«/mi»«mi»e«/mi»«msub»«mi»F«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»0255«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«mi»F«/mi»«mi»e«/mi»«msub»«mi»F«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»112«/mn»«mo».«/mo»«mn»85«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»m«/mi»«mrow»«mi»F«/mi»«mi»e«/mi»«msub»«mi»F«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»F«/mi»«mi»e«/mi»«msub»«mi»F«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»0255«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«mi»F«/mi»«mi»e«/mi»«msub»«mi»F«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»112«/mn»«mo».«/mo»«mn»85«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
Identify the correct formula and rearrange if necessary.
«math»«msub»«mi»m«/mi»«mrow»«mi»F«/mi»«mi»e«/mi»«msub»«mi»F«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mi»n«/mi»«mo»§#215;«/mo»«mi»M«/mi»«/math»
«math»«msub»«mi»m«/mi»«mrow»«mi»F«/mi»«mi»e«/mi»«msub»«mi»F«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mi»n«/mi»«mo»§#215;«/mo»«mi»M«/mi»«/math»
Substitute the values into the formula.
«math»«msub»«mi»m«/mi»«mrow»«mi»F«/mi»«mi»e«/mi»«msub»«mi»F«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»0255«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«mo»§#215;«/mo»«mn»112«/mn»«mo».«/mo»«mn»85«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/math»
«math»«msub»«mi»m«/mi»«mrow»«mi»F«/mi»«mi»e«/mi»«msub»«mi»F«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»0255«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«mo»§#215;«/mo»«mn»112«/mn»«mo».«/mo»«mn»85«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/math»
Calculate the answer.
«math»«msub»«mi»m«/mi»«mrow»«mi»F«/mi»«mi»e«/mi»«msub»«mi»F«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»2«/mn»«mo».«/mo»«mn»877«/mn»«mo»§#8230;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»=«/mo»«mn»2«/mn»«mo».«/mo»«mn»88«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#160;«/mo»«mo»(«/mo»«mi»to«/mi»«mo»§#160;«/mo»«mi»three«/mi»«mo»§#160;«/mo»«mi»significant«/mi»«mo»§#160;«/mo»«mi»digits«/mi»«mo»)«/mo»«/math»
«math»«msub»«mi»m«/mi»«mrow»«mi»F«/mi»«mi»e«/mi»«msub»«mi»F«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»2«/mn»«mo».«/mo»«mn»877«/mn»«mo»§#8230;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»=«/mo»«mn»2«/mn»«mo».«/mo»«mn»88«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#160;«/mo»«mo»(«/mo»«mi»to«/mi»«mo»§#160;«/mo»«mi»three«/mi»«mo»§#160;«/mo»«mi»significant«/mi»«mo»§#160;«/mo»«mi»digits«/mi»«mo»)«/mo»«/math»
What is the amount in moles of a 17 g sample of nickel(III) sulfate?
Write the chemical formula for the substance and determine its molar mass.
Ni3+ SO42–
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«mi»N«/mi»«msub»«mi»i«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»(«/mo»«mn»2«/mn»«mo»§#215;«/mo»«mn»58«/mn»«mo».«/mo»«mn»69«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mo»+«/mo»«mo»(«/mo»«mn»3«/mn»«mo»§#215;«/mo»«mn»32«/mn»«mo».«/mo»«mn»07«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mo»+«/mo»«mo»(«/mo»«mn»12«/mn»«mo»§#215;«/mo»«mn»16«/mn»«mo».«/mo»«mn»00«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«mi»N«/mi»«msub»«mi»i«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»405«/mn»«mo».«/mo»«mn»59«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«mtd»«/mtd»«mtd»«/mtd»«/mtr»«/mtable»«/math»
Ni3+ SO42–
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«mi»N«/mi»«msub»«mi»i«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»(«/mo»«mn»2«/mn»«mo»§#215;«/mo»«mn»58«/mn»«mo».«/mo»«mn»69«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mo»+«/mo»«mo»(«/mo»«mn»3«/mn»«mo»§#215;«/mo»«mn»32«/mn»«mo».«/mo»«mn»07«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mo»+«/mo»«mo»(«/mo»«mn»12«/mn»«mo»§#215;«/mo»«mn»16«/mn»«mo».«/mo»«mn»00«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«mi»N«/mi»«msub»«mi»i«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»405«/mn»«mo».«/mo»«mn»59«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«mtd»«/mtd»«mtd»«/mtd»«/mtr»«/mtable»«/math»
Liste the variables.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»N«/mi»«msub»«mi»i«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»m«/mi»«mrow»«mi»N«/mi»«msub»«mi»i«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»17«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«mi»N«/mi»«msub»«mi»i«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»405«/mn»«mo».«/mo»«mn»59«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»N«/mi»«msub»«mi»i«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»m«/mi»«mrow»«mi»N«/mi»«msub»«mi»i«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»17«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«mi»N«/mi»«msub»«mi»i«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»405«/mn»«mo».«/mo»«mn»59«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»
Identify the correct formula and rearrange if necessary.
«math»«msub»«mi»n«/mi»«mrow»«mi»N«/mi»«msub»«mi»i«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mfrac»«mi»m«/mi»«mi»M«/mi»«/mfrac»«/math»
«math»«msub»«mi»n«/mi»«mrow»«mi»N«/mi»«msub»«mi»i«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mfrac»«mi»m«/mi»«mi»M«/mi»«/mfrac»«/math»
Substitute the values into the formula.
«math»«msub»«mi»n«/mi»«mrow»«mi»N«/mi»«msub»«mi»i«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mfrac»«mrow»«mn»17«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«mrow»«mn»405«/mn»«mo».«/mo»«mn»59«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfrac»«/math»
«math»«msub»«mi»n«/mi»«mrow»«mi»N«/mi»«msub»«mi»i«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mfrac»«mrow»«mn»17«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«mrow»«mn»405«/mn»«mo».«/mo»«mn»59«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfrac»«/math»
Calculate the answer.
«math»«msub»«mi»n«/mi»«mrow»«mi»N«/mi»«msub»«mi»i«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»0419«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«mo»§#8230;«/mo»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»042«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«mo»§#160;«/mo»«mo»(«/mo»«mi»to«/mi»«mo»§#160;«/mo»«mi»two«/mi»«mo»§#160;«/mo»«mi»significant«/mi»«mo»§#160;«/mo»«mi»digits«/mi»«mo»)«/mo»«/math»
«math»«msub»«mi»n«/mi»«mrow»«mi»N«/mi»«msub»«mi»i«/mi»«mn»2«/mn»«/msub»«mo»(«/mo»«mi»S«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»0419«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«mo»§#8230;«/mo»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»042«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«mo»§#160;«/mo»«mo»(«/mo»«mi»to«/mi»«mo»§#160;«/mo»«mi»two«/mi»«mo»§#160;«/mo»«mi»significant«/mi»«mo»§#160;«/mo»«mi»digits«/mi»«mo»)«/mo»«/math»
Read This
Please read pages 108 and 109 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on the steps for calculating molar
mass. Remember, if you have any questions or you do not understand something, ask your teacher! Practice Questions
Complete the following practice question to check your understanding of the concept you just learned. Make sure you write a complete answer to the practice
question in your notes. After you have checked your answer, make corrections to your response (where necessary) to study from.- What is the amount in moles of a 125 g sample of copper(II) phosphate?
Write the chemical formula for the substance and determine its molar mass.
Cu2+ PO43–
«math»«msub»«mi»M«/mi»«mrow»«mi»C«/mi»«msub»«mi»u«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»P«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mo»(«/mo»«mn»3«/mn»«mo»§#215;«/mo»«mn»63«/mn»«mo».«/mo»«mn»55«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mo»+«/mo»«mo»(«/mo»«mn»2«/mn»«mo»§#215;«/mo»«mn»30«/mn»«mo».«/mo»«mn»97«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mo»+«/mo»«mo»(«/mo»«mn»8«/mn»«mo»§#215;«/mo»«mn»16«/mn»«mo».«/mo»«mn»00«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»M«/mi»«mrow»«mi»C«/mi»«msub»«mi»u«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»P«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»380«/mn»«mo».«/mo»«mn»59«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/math»List the variables.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«msub»«mi»u«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»P«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»m«/mi»«mrow»«mi»C«/mi»«msub»«mi»u«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»P«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»125«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«mi»C«/mi»«msub»«mi»u«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»P«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»380«/mn»«mo».«/mo»«mn»59«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»Identify the correct formula and rearrange if necessary.
«math»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«msub»«mi»u«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»P«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mfrac»«mi»m«/mi»«mi»M«/mi»«/mfrac»«/math»Substitute the values into the formula.
«math»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«msub»«mi»u«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»P«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mfrac»«mrow»«mn»125«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«mrow»«mn»380«/mn»«mo».«/mo»«mn»59«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfrac»«/math»Calcualte the answer.
«math»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«msub»«mi»u«/mi»«mn»3«/mn»«/msub»«mo»(«/mo»«mi»P«/mi»«msub»«mi»O«/mi»«mn»4«/mn»«/msub»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»3284«/mn»«mo»§#8230;«/mo»«mi»mol«/mi»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»328«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«mo»§#160;«/mo»«mo»(«/mo»«mi»to«/mi»«mo»§#160;«/mo»«mi»three«/mi»«mo»§#160;«/mo»«mi»significant«/mi»«mo»§#160;«/mo»«mi»digits«/mi»«mo»)«/mo»«/math» - What is the mass in grams of a 0.25 mol sample of calcium hydroxide?
Write the chemical formula for the substance and determine its molar mass.
Ca2+ OH–
«math»«msub»«mi»M«/mi»«mrow»«mi»C«/mi»«mi»a«/mi»«mo»(«/mo»«mi»O«/mi»«mi»H«/mi»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mo»(«/mo»«mn»1«/mn»«mo»§#215;«/mo»«mn»40«/mn»«mo».«/mo»«mn»08«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mo»+«/mo»«mo»(«/mo»«mn»2«/mn»«mo»§#215;«/mo»«mn»16«/mn»«mo».«/mo»«mn»00«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mo»+«/mo»«mo»(«/mo»«mn»2«/mn»«mo»§#215;«/mo»«mn»1«/mn»«mo».«/mo»«mn»01«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«mo»)«/mo»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi mathvariant=¨normal¨»M«/mi»«mrow»«mi»Ca«/mi»«mo»(«/mo»«mi»OH«/mi»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»74«/mn»«mo».«/mo»«mn»10«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/math»List the variables.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»m«/mi»«mrow»«mi»C«/mi»«mi»a«/mi»«mo»(«/mo»«mi»O«/mi»«mi»H«/mi»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»C«/mi»«mi»a«/mi»«mo»(«/mo»«mi»O«/mi»«mi»H«/mi»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»25«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«mi»C«/mi»«mi»a«/mi»«mo»(«/mo»«mi»O«/mi»«mi»H«/mi»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»74«/mn»«mo».«/mo»«mn»10«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»Identify the correct formula and rearrange if necessary.
«math»«msub»«mi»m«/mi»«mrow»«mi»C«/mi»«mi»a«/mi»«mo»(«/mo»«mi»O«/mi»«mi»H«/mi»«msub»«mo»)«/mo»«mn»2«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mi»n«/mi»«mo»§#215;«/mo»«mi»M«/mi»«/math»Substitute the values into the formula.
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The Law of Conservation of Mass
The total amount of reactants must equal the total amount of products.
You have already applied the law of conservation of mass when you balanced chemical reaction equations. The total number of atoms of each element needed to be carefully inspected and coefficients were added to ensure that the number of atoms of each
element on the reactant side of the equation equaled the number of atoms of each element on the product side.
Not only can you use the number of atoms to explain the law of conservation of mass, but you can also use the number of moles or even the mass. Using the combustion of methane as our example, this chart compares the amount of reactants to the amount of products.
Not only can you use the number of atoms to explain the law of conservation of mass, but you can also use the number of moles or even the mass. Using the combustion of methane as our example, this chart compares the amount of reactants to the amount of products.
| Balanced Chemical Equation |
CH4(g) + |
2O2(g) | → | CO2(g) + |
2H2O(g) |
| Number of Particles | 1 molecule of CH4 | 2 molecules of O2 |
→ | 1 molecule of CO2 | 2 molecules of H2O |
| Amount (mol) | 1 mol CH4 | 2 mol O2 | → | 1 mol CO2 | 2 mol H2O |
| Mass (g) |
16.05 g |
64.00 g |
→ | 44.01 g |
36.04 g |
| Total Mass (g) |
80.05 g |
→ | 80.05 g |
||
Read This
Please read page 111 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on applying the law of conservation of mass
to the mole concept. Remember, if you have any questions or you do not understand something, ask your teacher! Practice Questions
Complete the following practice question to check your understanding of the concept you just learned. Make sure you write a complete answer to the practice
question in your notes. After you have checked your answer, make corrections to your response (where necessary) to study from.- Predict the mass of nitrogen monoxide gas that should be produced when 28.02 g of nitrogen gas reacts with 32.0 g of oxygen gas.
According to the law of conservation of mass, the amount of nitrogen monoxide produced should be 60.02 g.
Conclusion
The mole provides a convenient way to convert between the number of atoms and the mass of a substance.
B13.10 A scientist weighing a mixture
The mole is a method of counting; one mole is equal to 6.02 × 1023 particles. Such a large number of particles of a substance is required because atoms and molecules are so small.
The molar mass of a substance is a way of converting between moles and mass. By knowing the mass of a sample, molar mass can be easily measured using a laboratory scale.
You also looked at the law of conservation of mass, which states that during a chemical reaction, matter is not created or destroyed but is converted from one form to another.
The molar mass of a substance is a way of converting between moles and mass. By knowing the mass of a sample, molar mass can be easily measured using a laboratory scale.
You also looked at the law of conservation of mass, which states that during a chemical reaction, matter is not created or destroyed but is converted from one form to another.
Watch This
Introduction to Moles @YouTube Tyler DeWitt https://adlc.wistia.com/medias/z1iz2etcrd
Watch this video to review the concept of the mole. Counting Atoms: Intro to Moles Part 2 @ YouTube Tyler DeWitt
Watch this video to review the concept of molar mass.Converting between Grams and Moles @YouTube Tyler DeWitt
Watch this video to review the concept of converting between mass and moles.
2.6 Assignment
Unit 2 Assignment Lessons 10-13
It is now time to complete the Lesson 13 portion of 2.6 Assignment. This assignment has two parts.
- Part 1 Written Portion: Select the preferred document type from the options below. Download and save the assignment on your desktop (or documents folder).
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- Once you have completed the written portion of your assignment, click on the button below to go to the submission page.
Written Portion Submission Page - Part 2 Online Portion: It is now time to complete the Lesson 5 questions of the online portion of this assignment. Click on the button below to go to the online questions of this assignment.
Online Questions
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