Key Concepts
⚛ 1 mole of a pure substance has a mass equal to its molecular mass^{(1)} expressed in grams.
· This is known as the molar mass, M, and has the units g mol^{-1}
(g/mol, grams per mole of substance)
⚛ The relationship between molar mass, mass and moles can be expressed as a mathematical equation as shown below:
g mol^{-1} = g ÷ mol
molar mass = mass ÷ moles
M = m ÷ n
where:
M = molar mass of the pure substance (measured in g mol^{-1})
m = mass of the pure substance (measured in grams, g)
n = amount of the pure substance (measured in moles, mol)
⚛ This mathematical equation can be rearranged to give the following:
(i) n = m ÷ M
moles = mass ÷ molar mass
(ii) m = n × M
mass = moles × molar mass
⚛ To calculate the moles of pure substance: n = m ÷ M
⚛ To calculate mass of pure substance: m = n × M
⚛ To calculate molar mass of pure substance: M = m ÷ n
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Calculating the Mass of a Pure Substance (m=nM)
1 mole of a pure substance is defined as having a mass in grams equal to its relative molecular mass.
This quantity is known as the molar mass (symbol M).
So, mass of 1 mole of a pure substance = relative molecular mass in grams
And, mass of 1 mole of a pure substance = molar mass of the pure substance (g mol^{-1})
Or, mass of 1 mole = M (g mol^{-1})
The table below gives the mass of 1 mole of a number of common pure substances:
name | molecularformula | relative molecular mass | molar mass (g mol^{-1}) | mass of 1 mole (g) |
---|---|---|---|---|
helium gas | He | 4.003 | 4.003 g mol^{-1} | 4.003 g |
oxygen gas | O_{2} | 2 × 16.00 = 32.00 | 32.00 g mol^{-1} | 32.00 g |
carbon dioxide gas | CO_{2} | 12.01 + (2 × 16.00) = 44.01 | 44.01 g mol^{-1} | 44.01 g |
liquid water | H_{2}O | (2 × 1.008) + 16.00 = 18.016 | 18.016 g mol^{-1} | 18.016 g |
From the table we see that 1 mole of water has a mass of 18.016 grams, which isn't very much (about the mass of water in a couple of small ice-cubes you'd make in your family freezer).
But what if you had 10 moles of water? What would be the mass of 10 moles of water?
If 1 mole of water has a mass of 18.016 g, then 10 moles of water must have ten times more mass:
mass of 10 moles of water = 10 × mass of 1 mole of water
mass of 10 moles of water = 10 × 18.016 = 180.16 g (about the mass of water you could put in a small glass)
So, if we only had ½ mole of water, what mass of water would we have?
If 1 mole of water has a mass of 18.016 g, then ½ mole of water must have ½ the mass:
mass of ½ mole of water = ½ × mass of 1 mole of water
mass of ½ mole of water = ½ × 18.016 = 9.008 g
In both of the examples above, we can calculate the mass of water in grams by multiplying the moles of water by the mass of 1 mole of water in grams:
mass water = moles of water × mass of 1 mole water
because the mass of 1 mole of water in grams is known as its molar mass, we can write:
mass water = moles of water × molar mass of water
The table below compares the mass of different amounts of water in moles:
mass of water (g) | = | moles of water (mol) | × | mass of 1 mole of water (molar mass of water) (g mol^{-1}) |
---|---|---|---|---|
0 | = | 0.00 | × | 18.016 |
9.008 | = | 0.50 | × | 18.016 |
18.016 | = | 1.00 | × | 18.016 |
27.024 | = | 1.50 | × | 18.016 |
180.16 | = | 10.00 | × | 18.016 |
270.24 | = | 15.00 | × | 18.016 |
From the data in the table we can generalise and say that for any pure substance the mass of substance in grams is equal to the moles of substance multiplied by the mass of 1 mole of the substance:
mass = moles × mass of 1 mole
and since mass of 1 mole of a substance (in grams) = molar mass (in grams per mole)
mass (g) = moles × molar mass (g mol^{-1})
m = n × M
where
m = mass of pure substance in grams
n = amount of pure substance in moles
M = molar mass of pure substance in grams per mole
We could also plot the data in the table above on a graph as shown below:
mass (g) | |
moles (mol) |
This graph above shows a straight line that passes through the origin (0,0) so the equation for the line is:
y = slope × x
where:
y is mass of water (g)
x is moles of water (mol)
slope (gradient) of the line = vertical rise ÷ horizontal run
We can determine the slope of the line using 2 points on the straight line, for example, (0,0) and (15.0, 270.24):
slope = (270.24 g - 0 g) ÷ (15 mol - 0 mol) = 18.016 g mol^{-1}
Since 18.016 g mol^{-1} is the molar mass of water, we can say:
slope = molar mass of water (g mol^{-1})
Therefore the equation for this line is:
mass (H_{2}O) = molar mass (H_{2}O) × moles (H_{2}O)
In general:
mass (g) = molar mass (g mol^{-1}) × moles (mol)
From the data in the table and its graphical representation, we can generalise and say that for any pure substance the mass of substance in grams is equal to the moles of substance multiplied by the mass of 1 mole of the substance:
mass = moles × mass of 1 mole
and since mass of 1 mole of a substance (in grams) = molar mass (in grams per mole)
mass (g) = moles × molar mass (g mol^{-1})
m = n × M
Follow these steps to calculate the mass of a pure substance given the amount of substance in moles:
Step 1. Extract the data from the question:
mass = m = ? (units are grams)
moles = n = write down what you are told in the question
molar mass = M = write down what you are told in the question (units are g mol^{-1})
(you may need to calculate this using the molecular formula of the pure substance and a Periodic Table)
Step 2. Check the units for consistency and convert if necessary:
The amount of substance must be in moles (mol) !
If amount is given in millimoles (mmol), divide it by 1,000 to give the amount in moles (mol).
If amount is given in micromoles (μmol), divide it by 1,000,000 to give an amount in moles (mol).
If amount is given in kilomoles (kmol), multiply it by 1,000 to give an amount in moles (mol).
Step 3. Write the mathematical equation (mathematical formula):
mass = moles × molar mass
or
m = n × M
Step 4. Substitute in the values and solve the equation to find the value of mass, m, in grams (g).
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Calculating the Moles of a Pure Substance (n=m/M)
In the discussion above, we discovered that we could calculate the mass of a pure substance using the moles and molar mass of the substance:
mass (g) = moles (mol) × molar mass (g mol^{-1})
How would we calculate the moles of pure substance if we knew the mass of the substance?
(a) We could use some algebra: divide both sides of the equation by the molar mass:
mass | = | moles × |
molar mass | |
moles = mass ÷ molar mass
n = m ÷ M
or
(b) We could use some logic:
we know the mass with units of | grams (g) |
we know the molar mass with units of | grams per mole (g mol^{-1}) |
we need to find moles with units of | mole (mol) |
By inspection of units we see that dividing molar mass by mass will give us a quantitiy in units of "mol^{-1}"
molar mass/mass = g mol^{-1}/g = mol^{-1}
If we turn this upside down (in mathematical terms, take the reciprocal) we get a quantity with units of "mol" which is what we want:
mass/molar mass = g/g mol^{-1} = mol
so
moles = mass ÷ molar mass
n = m ÷ M
Follow these steps to calculate the amount of pure substance in moles given the mass of substance:
Step 1. Extract the data from the question:
mass = m = write down what you are told in the question
moles = n = ? (units are mol)
molar mass = M = write down what you are told in the question (units are g mol^{-1})
(you may need to calculate this using the molecular formula of the pure substance and a Periodic Table)
Step 2. Check the units for consistency and convert if necessary:
Mass must be in grams !
If mass is given in milligrams (mg), divide it by 1,000 to give the mass in grams (g).
If mass is given in micrograms (μg), divide it by 1,000,000 to give a mass in grams (g).
If mass is given in kilograms (kg), multiply it by 1,000 to give a mass in grams (g).
Step 3. Write the mathematical equation (mathematical formula):
moles = mass ÷ molar mass
or
n = m ÷ M
Step 4. Substitute in the values and solve the equation to find moles of substance (mol).
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Calculating the Molar Mass of a Pure Substance (M=m/n)
What if you knew the amount of a pure substance in moles and its mass?
Could you calculate its molar mass?
Recall that mass = moles × molar mass or m = n × M
(a) We could use some algebra: divide both sides of the equation by the moles:
mass | = | |
moles | |
molar mass = mass ÷ moles
M = m ÷ n
or (b) We could use some logic:
By inspection of units we see that dividing the mass in grams by the amount in moles we arrive at a quantity with the units grams per mole (g mol^{-1}) which are the units for molar mass.
Therefore, molar mass (g mol^{-1}) = mass (g) ÷ moles (mol)
or you can write
M = m ÷ n
Follow these steps to calculate the molar mass of a pure substance given the amount of substance in moles and the mass of substance:
Step 1. Extract the data from the question:
mass = m = write down what you are told in the question
moles = n = write down what you are told in the question
molar mass = M = ? (units are g mol^{-1})
Step 2. Check the units for consistency and convert if necessary:
Mass must be in grams (g)!
Amount, moles, must be in moles (mol)!
Step 3. Write the mathematical equation (mathematical formula):
molar mass = mass ÷ moles
or
M = m ÷ n
Step 4. Substitute in the values and solve the equation to find the molar mass of the substance in grams per mole.
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Worked Examples of Calculating Mass, Moles, Molar Mass
In each of the worked examples below, you will be asked to calculate either the moles, mass, or molar mass of a pure substance.
To answer each question correctly you will need to:
- Step 1. Extract the information from the question
- Step 2. Check the data for consistency of units
- Step 3. Choose a mathematical equation to find the unknown
- Step 4. Substitute your values into the equation and solve
Worked Example: mass = moles × molar mass (m=n×M)
Question: Calculate the mass of 0.25 moles of water, H_{2}O.
Solution:
Step 1. Extract the data from the question:
moles = n = 0.25 mol
molar mass = M = (2 × 1.008) + 16.00 = 18.016 g mol^{-1}
(Calculated using the periodic table)
mass = m = ? g
Step 2. Check the data for consistency:
Is the amount of water in moles (mol)? Yes.
We do not need to convert this.
Step 3. Write the mathematical equation (mathematical formula):
mass = moles × molar mass
or
m = n × M
Step 4. Substitute the values into the equation and solve for mass (g):
mass = m = 0.25 mol × 18.016 g mol^{-1} = 4.5 g
(only 2 significant figures are justified)
Worked Example: moles = mass ÷ molar mass (n=m/M)
Question: Calculate the amount of oxygen gas, O_{2}, in moles present in 124.5 g of oxygen gas.
Solution:
Step 1. Extract the data from the question:
mass = m = 124.5 g
molar mass = M = 2 × 16.00 =32.00 g mol^{-1}
(Calculated using the periodic table)
moles = n = ? mol
Step 2. Check the data for consistency:
Is the mass of oxygen gas in grams (g)? Yes.
We do not need to convert this.
Step 3. Write the mathematical equation (mathematical formula):
moles = mass ÷ molar mass
or
n = m ÷ M
Step 4. Substitute the values into the equation and solve to find moles of oxygen gas:
moles = n = 124.5 g ÷ 32.00 g mol^{-1} = 3.891 mol
(4 significant figures are justified)
Worked Example: molar mass = mass ÷ moles (M=m/n)
Question: Calculate the molar mass of a pure substance if 1.75 moles of the substance has a mass of 29.79 g.
Solution:
Step 1. Extract the data from the question:
mass = m = 29.79 g
moles = n = 1.75 mol
Step 2. Check the data for consistency:
Is the mass of in grams (g)? Yes. We do not need to convert this.
Is the amount of substance in moles (mol)? Yes. We do not need to convert this.
Step 3. Write the equation:
molar mass = mass ÷ moles
or
M = m ÷ n
Step 4. Substitute the values into the equation and solve for molar mass:
molar mass = M = 29.79 g ÷ 1.75 mol = 17.0 g mol^{-1}
(3 significant figures are justified)
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Problem Solving Using Moles, Mass, and Molar Mass
The Problem: Calcium carbonate, CaCO_{3}, is an important industrial chemical.
Chris the Chemist has an impure sample of calcium carbonate.The mass of the impure sample is 0.1250 kg and it is composed of 87.00% (by mass) calcium carbonate.Before Chris can use this calcium carbonate in a chemical reaction, Chris needs to know the amount, in moles, of calcium carbonate present in this sample.
Calculate the amount of calcium carbonate in moles present in this impure sample of calcium carbonate.
Solving the Problem using the StoPGoPS model for problem solving:
State the question. | What is the question asking you to do? Calculate the amount of calcium carbonate in moles n(CaCO_{3}) = moles of calcium carbonate = ? mol | |
Pause to Plan. | What chemical principle will you need to apply? Apply stoichoimetry (n = m ÷ M) What information (data) have you been given?
What is your plan for solving this problem? (i) Write the mathematical equation to calculate moles of calcium carbonate: n(mol) = m(g) ÷ M(g mol^{-1}) (ii) Calculate the mass of calcium carbonate in the sample in kilograms (kg). mass of calcium carbonate (kg) = 87.00% of mass of sample (kg) m(CaCO_{3}) in kilograms = (87.00/100) × m(sample) (iii) Convert the mass of calcium carbonate in kilograms (kg) to mass in grams (g) m(CaCO_{3}) in grams = m(CaCO_{3}) in kg × 1000 g/kg (iv) Calculate the molar mass of calcium carbonate molar mass = M(CaCO_{3}) = M(Ca) + M(C) + [3 × M(O)] = (v) Substitute the values for m(CaCO_{3}) in grams and M(CaCO_{3}) in g mol^{-1} into the mathematical equation and solve for n (mol) n(mol) = m(g) ÷ M(g mol^{-1}) | |
Go with the Plan. | (i) Write the mathematical equation to calculate moles of calcium carbonate: n(mol) = m(g) ÷ M(g mol^{-1}) n(CaCO_{3}) = m(CaCO_{3}) ÷ M(CaCO_{3}) (ii) Calculate the mass of calcium carbonate in the sample in kilograms (kg). mass of calcium carbonate (kg) = 87.00% of mass of sample (kg) m(CaCO_{3}) in kilograms = (87.00/100) × m(sample) m(CaCO_{3}) in kilograms = (87.00/100) × 0.1250 kg = 0.10875 kg (iii) Convert the mass of calcium carbonate in kilograms (kg) to mass in grams (g) m(CaCO_{3}) in grams = m(CaCO_{3}) in kg × 1000 g/kg m(CaCO_{3}) in grams = 0.10875 kg × 1000 g/kg = 108.75 g (iv) Calculate the molar mass of calcium carbonate molar mass = M(CaCO_{3}) = M(Ca) + M(C) + [3 × M(O)] M(CaCO_{3}) = 40.08 + 12.01 + [3 × 16.00] = 40.08 + 12.01 + 48.00 = 100.09 g mol^{-1} (v) Substitute the values for m(CaCO_{3}) in grams and M(CaCO_{3}) in g mol^{-1} into the mathematical equation and solve for n (mol) n(CaCO_{3}) = m(CaCO_{3}) ÷ M(CaCO_{3}) n(CaCO_{3}) = m(CaCO_{3}) ÷ M(CaCO_{3}) = 108.75 (4 significant figures are justified) | |
Ponder Plausability. | Have you answered the question that was asked? Yes, we have calculated the moles of calcium carbonate in the sample. Is your solution to the question reasonable? Let's work backwards to see if the moles of calcium carbonate we have calculated will give us the correct mass for the sample. Since this approximate value for the mass of the sample is about the same as the mass of sample given in the question, we are reasonably confident that our answer is correct. | |
State the solution. | How many moles of calcium carbonate are present in the sample? n(CaCO_{3}) = 1.087 mol |
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Sample Question: Moles, Mass, and Molar Mass
Determine the mass in grams of 0.372 moles of solid rhombic sulfur (S_{8}).
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Footnotes:
(1) Molecular mass is also known as molecular weight, formula weight or formula mass
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