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Calculating percentage mass
1. Calculate the percentage by mass of iron in iron(III) oxide, Fe2O3.
Relative atomic masses, Ar: Fe = 56 O = 16
( 2 x 56 ) / ( (2 x 56) + (3 x 16) )
112 / 160
0.7 x 100 = 70 %
2. A compound contains 10 g of hydrogen and 80 g of oxygen.
What is the empirical formula of this compound?
Element H O
Mass in g 10 g 80 g
Ar 1 g/mol 16 g/mol
Number of 10 5
moles
Molar ratio 10/5 5/5
Molar ratio 2 1
Empirical formula : H2O
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Now solve this one in the same way:
Substance X was analyzed and found to contain 31.58% carbon, 5.26% hydrogen and
63.16% oxygen by mass.
- What is the empirical formula of substance X?
Relative atomic masses, Ar: C = 12; H = 1; O = 16
3. Calculating Molecular Formula
Molecular formula gives the actual numbers of atoms of each element present in
the formula of the compound
To calculate the molecular formula:
Step 1. Find the relative formula mass of the empirical formula
Step 2. Use the following equation:
Step 3. Multiply the number of each element present in the empirical formula by the
number from step 2 to nd the molecular formula
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- The empirical formula of X is C4H10S1
The relative formula mass (Mr ) of X is 180.
What is the molecular formula of X?
( Relative atomic mass, Ar: Carbon : 12 Hydrogen : 1 Sulfur : 32 )
4. Deducing formulae of hydrated salts
1. The formula of hydrated salts can be determined experimentally by weighing a sample
of the hydrated salt, heating it until the water of crystallization has been driven o , then
reweighing the now anhydrous salt.
2. From the results, you can determine the mass of anhydrous salt and the mass of the
water of crystallization.
3. Applying a similar approach to deducing empirical formulae, the formula of the
hydrated salt can be calculated
- 11.25 g of hydrated copper sulfate, CuSO4.XH2O, is heated until it loses all of its water
of crystallization. It is reweighed and its mass is 7.19 g. What is the formula of the
hydrated copper(II) sulfate?
Compound CuSO4 H2O
Mass 7.19 g (11.25-7.19) 4.06 g
Mr 160 g/mol 18 g/ mol
n. Moles 0.045 0.226
Molar ratio 0.045 / 0.045 0.226 /0.045
1 5
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So the formula is CuSO4.5H2O
5. In a reaction to produce sodium sul de, Na2S, 9.2 g of sodium is reacted with 8.0 g of
sulfur. Which reactant is in excess and which is limiting?
- Write the balanced equation
2Na + S —> Na2S
Reactant Na S
Mass 9.2 g 8.0 g
Ar 23 g/ mol 32 g/ mol
N. Moles 0.4 0.25
Divide by 0.2 0.25
coe cient
0.2 < 0.25 so sodium is the limiting reactant.
- To nd the amount of remaining excess reactant, subtract the mass of excess
reactant consumed during the reaction from the total mass of excess reactant given
in the question.
To nd the amount consumed use molar ratio:
2 mol of Na react with 1 mol of S
0.4 mol of Na react with (X) mol of S
Cross product : 0.4 = 2 X X = 0.2
Mass of sulfur = n.moles x Ar = 0.2 mol * 32 g/mol = 6.4 g
The amount of remaining excess reactant = 8.0 - 6.4 = 1.6 g
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6. 64 g of methanol, CH3OH, reacts with 96 g of oxygen gas to produce 88 g of carbon
dioxide and 72 g of water.
Deduce the balanced equation for the reaction.
(C = 12, H = 1, O = 16)
- Just nd the number of moles ( mass / molar mass ), remember that it is H2
( diatomic molecule).
- Write the chemical equation putting the number of moles as coe cients.
7. Calculate the mass of magnesium oxide that can be made by completely burning 6.0 g
of magnesium in oxygen in the following reaction:
2Mg (s) + O2 (g) ⟶ 2MgO (s)
Use the molar ratio ( cross product) after converting mass to moles to nd the number of
moles of MgO then multiply by Mr to nd the mass in g.
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8. What is the mass of 0.25 moles of zinc?
9. How many moles are in 2.64 g of sucrose, C12H22O11 (Mr = 342.3)?
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