Mole Calculations 4
1. How many potassium
atoms are in 272 grams of
potassium chloride?
𝟐. 𝟐𝟎 × 𝟏𝟎𝟐𝟒 𝒂𝒕𝒐𝒎𝒔
2. How many magnesium
atoms are in 357 grams of
magnesium oxide?
𝟓. 𝟑𝟑 × 𝟏𝟎𝟐𝟒 𝒂𝒕𝒐𝒎𝒔
3. How many oxygen atoms
are in 463 grams of calcium
hydroxide?
𝟕. 𝟓𝟑 × 𝟏𝟎𝟐𝟒 𝒂𝒕𝒐𝒎𝒔
4. How many sodium
atoms are in 521 grams of
sodium carbonate?
𝟓. 𝟗𝟐 × 𝟏𝟎𝟐𝟒 𝒂𝒕𝒐𝒎𝒔
5. How many grams of
water would be produced
from burning 29.0 grams of
hydrogen gas?
𝟐𝟓𝟗𝒈
6. How many grams of
water would be produced
from burning 134.0 grams
of hydrogen gas?
𝟏𝟏𝟗𝟕𝒈
Mole Calculations 4 Solutions
Step 1: Write the chemical formula of the given substance.
KCl
Step 2: Calculate the number of moles of the given substance.
𝑚𝐾𝐶𝑙
272𝑔
𝑛𝐾𝐶𝑙 =
=
= 3.648558 𝑚𝑜𝑙
𝑀𝐾𝐶𝑙 (39.10 + 35.45)𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 3: Calculate the number of moles of the specified element.
From the chemical formula: 𝑛𝐾 = 𝑛𝐾𝐶𝑙 = 3.648558 𝑚𝑜𝑙
Step 4: Calculate the number of atoms of the specified element.
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑡𝑎𝑠𝑠𝑖𝑢𝑚 𝑎𝑡𝑜𝑚𝑠 = 𝑛𝐾 × 𝑁𝐴 = 3.648558 × 6.022 × 1023 = 2.20 × 1024
Step 1: Write the chemical formula of the given substance.
MgO
Step 2: Calculate the number of moles of the given substance.
𝑚𝑀𝑔𝑂
357𝑔
𝑛𝑀𝑔𝑂 =
=
= 8.85636 𝑚𝑜𝑙
𝑀𝑀𝑔𝑂 (24.31 + 16.00)𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 3: Calculate the number of moles of the specified element.
From the chemical formula: 𝑛𝑀𝑔 = 𝑛𝑀𝑔𝑂 = 8.85636 𝑚𝑜𝑙
Step 4: Calculate the number of atoms of the specified element.
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑎𝑔𝑛𝑒𝑠𝑖𝑢𝑚 𝑎𝑡𝑜𝑚𝑠 = 𝑛𝑀𝑔 × 𝑁𝐴 = 8.85636 × 6.022 × 1023 = 5.33 × 1024
Step 1: Write the chemical formula of the given substance.
Ca(OH)2
Step 2: Calculate the number of moles of the given substance.
𝑚𝐶𝑎(𝑂𝐻)2
463𝑔
𝑛𝐶𝑎(𝑂𝐻)2 =
=
= 6.24865 𝑚𝑜𝑙
𝑀𝐶𝑎(𝑂𝐻)2 (40.08 + (16.00 + 1.008) × 2)𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 3: Calculate the number of moles of the specified element.
From the chemical formula: 𝑛𝑂 = 2 × 𝑛𝐶𝑎(𝑂𝐻)2 = 12.49730 𝑚𝑜𝑙
Step 4: Calculate the number of atoms of the specified element.
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑥𝑦𝑔𝑒𝑛 𝑎𝑡𝑜𝑚𝑠 = 𝑛𝑂 × 𝑁𝐴 = 12.49730 × 6.022 × 1023 = 7.53 × 1024
Step 1: Write the chemical formula of the given substance.
Na2CO3
Step 2: Calculate the number of moles of the given substance.
𝑚𝑁𝑎2 𝐶𝑂3
521𝑔
𝑛𝑁𝑎2 𝐶𝑂3 =
=
= 4.91556 𝑚𝑜𝑙
𝑀𝑁𝑎2 𝐶𝑂3 (22.99 × 2 + 12.01 + 16.00 × 3)𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 3: Calculate the number of moles of the specified element.
From the chemical formula: 𝑛𝑁𝑎 = 2 × 𝑛𝑁𝑎2 𝐶𝑂3 = 9.83112 𝑚𝑜𝑙
Step 4: Calculate the number of atoms of the specified element.
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑜𝑑𝑖𝑢𝑚 𝑎𝑡𝑜𝑚𝑠 = 𝑛𝑁𝑎 × 𝑁𝐴 = 9.83112 × 6.022 × 1023 = 5.92 × 1024
Step 1: Write the balanced equation for the reaction.
2H2 + O2 → 2H2O
H2 + ½ O2 → H2O
Step 2: Calculate the number of moles of the given substance.
𝑚𝐻2
29.0𝑔
𝑛𝐻2 =
=
= 14.38492 𝑚𝑜𝑙
𝑀𝐻2 (1.008 × 2)𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 3: Calculate the number of moles of the specified substance.
From the balanced equation: 𝑛𝐻2 𝑂 = 𝑛𝐻2 = 14.38492 𝑚𝑜𝑙
Step 4: Calculate the mass of the specified substance.
𝑚𝐻2 𝑂 = 𝑛𝐻2 𝑂 × 𝑀𝐻2 𝑂 = 14.38492𝑚𝑜𝑙 × (1.008 × 2 + 16.00)𝑔 ⋅ 𝑚𝑜𝑙 −1 = 259𝑔
Step 1: Write the balanced equation for the reaction.
2H2 + O2 → 2H2O
H2 + ½ O2 → H2O
Step 2: Calculate the number of moles of the given substance.
𝑚𝐻2
134.0𝑔
𝑛𝐻2 =
=
= 66.46825 𝑚𝑜𝑙
𝑀𝐻2 (1.008 × 2)𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 3: Calculate the number of moles of the specified substance.
From the balanced equation: 𝑛𝐻2 𝑂 = 𝑛𝐻2 = 66.46825 𝑚𝑜𝑙
Step 4: Calculate the mass of the specified substance.
𝑚𝐻2 𝑂 = 𝑛𝐻2 𝑂 × 𝑀𝐻2 𝑂 = 66.46825𝑚𝑜𝑙 × (1.008 × 2 + 16.00)𝑔 ⋅ 𝑚𝑜𝑙 −1 = 1197𝑔
7. How many grams of
chlorine gas would be
produced from
decomposing 1.653kg of
sodium chloride?
𝟏𝟎𝟎𝟑𝒈
8. How many grams of
carbon dioxide gas would
be produced from
decomposing 5.345g of
copper (II) carbonate?
𝟏. 𝟗𝟎𝟒𝒈
9. How many grams of
water are needed to
produce 18.25 grams of
glucose in the
photosynthesis reaction?
𝟏𝟎. 𝟗𝟓𝒈
10. How many grams of
nitrogen gas are needed to
produce 27.00 grams of
ammonia in the Haber
process?
𝟐𝟐. 𝟐𝟏𝒈
11. How many kilograms of
oxygen are needed to
completely combust
152.0kg of benzene?
𝟒𝟔𝟕. 𝟎𝒌𝒈
12. How many kilograms of
oxygen are needed to
completely combust
92.15kg of ethane?
𝟑𝟒𝟑. 𝟐𝒌𝒈
Step 1: Write the balanced equation for the reaction.
2NaCl→ 2Na + Cl2
NaCl→ Na + ½ Cl2
Step 2: Calculate the number of moles of the given substance.
𝑚𝑁𝑎𝐶𝑙
1653𝑔
𝑛𝑁𝑎𝐶𝑙 =
=
= 28.28542 𝑚𝑜𝑙
𝑀𝑁𝑎𝐶𝑙 (22.99 + 35.45)𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 3: Calculate the number of moles of the specified substance.
From the balanced equation: 𝑛𝐶𝑙2 = 12𝑛𝑁𝑎𝐶𝑙 = 14.14271 𝑚𝑜𝑙
Step 4: Calculate the mass of the specified substance.
𝑚𝐶𝑙2 = 𝑛𝐶𝑙2 × 𝑀𝐶𝑙2 = 14.14271𝑚𝑜𝑙 × (35.45 × 2)𝑔 ⋅ 𝑚𝑜𝑙 −1 = 1003𝑔
Step 1: Write the balanced equation for the reaction.
CuCO3 → CuO + CO2
Step 2: Calculate the number of moles of the given substance.
𝑚𝐶𝑢𝐶𝑂3
5.345𝑔
𝑛𝐶𝑢𝐶𝑂3 =
=
= 0.043258 𝑚𝑜𝑙
𝑀𝐶𝑢𝐶𝑂3 (63.55 + 12.01 + 16.00 × 3)𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 3: Calculate the number of moles of the specified substance.
From the balanced equation: 𝑛𝐶𝑂2 = 𝑛𝐶𝑢𝐶𝑂3 = 0.043258 𝑚𝑜𝑙
Step 4: Calculate the mass of the specified substance.
𝑚𝐶𝑂2 = 𝑛𝐶𝑂2 × 𝑀𝐶𝑂2 = 0.043258𝑚𝑜𝑙 × (12.01 + 16.00 × 2)𝑔 ⋅ 𝑚𝑜𝑙 −1 = 1.904𝑔
Step 1: Write the balanced equation for the reaction.
6CO2 + 6H2O → C6H12O6 + 6O2
Step 2: Calculate the number of moles of the given substance.
𝑚𝑔𝑙𝑢𝑐𝑜𝑠𝑒
18.25𝑔
𝑛𝑔𝑙𝑢𝑐𝑜𝑠𝑒 =
=
= 0.10130 𝑚𝑜𝑙
𝑀𝑔𝑙𝑢𝑐𝑜𝑠𝑒 (12.01 × 6 + 1.008 × 12 + 16.00 × 6)𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 3: Calculate the number of moles of the specified substance.
From the balanced equation: 𝑛𝐻2 𝑂 = 6𝑛𝑔𝑙𝑢𝑐𝑜𝑠𝑒 = 0.60781 𝑚𝑜𝑙
Step 4: Calculate the mass of the specified substance.
𝑚𝐻2 𝑂 = 𝑛𝐻2 𝑂 × 𝑀𝐻2 𝑂 = 0.60781𝑚𝑜𝑙 × (1.008 × 2 + 16.00)𝑔 ⋅ 𝑚𝑜𝑙 −1 = 10.95𝑔
Step 1: Write the balanced equation for the reaction.
N2 + 3H2 → 2NH3
½ N2 + 32 H2 → NH3
Step 2: Calculate the number of moles of the given substance.
𝑚𝑁𝐻3
27.00𝑔
𝑛𝑁𝐻3 =
=
= 1.58507 𝑚𝑜𝑙
𝑀𝑁𝐻3 (14.01 + 1.008 × 3)𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 3: Calculate the number of moles of the specified substance.
From the balanced equation: 𝑛𝑁2 = 12 𝑛𝑁𝐻3 = 0.79253 𝑚𝑜𝑙
Step 4: Calculate the mass of the specified substance.
𝑚𝑁2 = 𝑛𝑁2 × 𝑀𝑁2 = 0.79253𝑚𝑜𝑙 × (14.01 × 2)𝑔 ⋅ 𝑚𝑜𝑙 −1 = 22.21𝑔
Step 1: Write the balanced equation for the reaction.
2C6H6 + 15O2 → 12CO2 + 6H2O
C6H6 + 152 O2 → 6CO2 + 3H2O
Step 2: Calculate the number of moles of the given substance.
𝑚𝐶6 𝐻6
152000𝑔
𝑛𝐶6 𝐻6 =
=
= 1946.023𝑚𝑜𝑙
𝑀𝐶6 𝐻6 (12.01 × 6 + 1.008 × 6)𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 3: Calculate the number of moles of the specified substance.
From the balanced equation: 𝑛𝑂2 = 15
𝑛
= 14595.1759 𝑚𝑜𝑙
2 𝐶6 𝐻6
Step 4: Calculate the mass of the specified substance.
𝑚𝑂2 = 𝑛𝑂2 × 𝑀𝑂2 = 14595.1759𝑚𝑜𝑙 × (16.00 × 2)𝑔 ⋅ 𝑚𝑜𝑙 −1 = 467046𝑔
Step 1: Write the balanced equation for the reaction.
2C2H6 +7O2 → 4CO2 + 6H2O
C2H6 + 72 O2 → 2CO2 + 3H2O
Step 2: Calculate the number of moles of the given substance.
𝑚𝐶2 𝐻6
92150𝑔
𝑛𝐶2 𝐻6 =
=
= 3064.71997𝑚𝑜𝑙
𝑀𝐶2 𝐻6 (12.01 × 2 + 1.008 × 6)𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 3: Calculate the number of moles of the specified substance.
From the balanced equation: 𝑛𝑂2 = 72𝑛𝐶2 𝐻6 = 10726.51989 𝑚𝑜𝑙
Step 4: Calculate the mass of the specified substance.
𝑚𝑂2 = 𝑛𝑂2 × 𝑀𝑂2 = 10726.51989𝑚𝑜𝑙 × (16.00 × 2)𝑔 ⋅ 𝑚𝑜𝑙 −1 = 343249𝑔
13. How many litres of CO2
measured at 25oC and
100kPa would be produced
from completely
combusting 235.5g of npentane?
𝟒𝟎𝟒. 𝟔𝑳
14. How many litres of CO2
measured at 25oC and
100kPa would be produced
from completely
combusting 94.78g of noctane?
𝟏𝟔𝟒. 𝟔𝑳
15. How many litres of H2
measured at 25oC and
100kPa would be produced
from completely reacting
15.00g of magnesium with
hydrochloric acid?
𝟏𝟓. 𝟑𝟎𝑳
16. How many litres of H2
measured at 25oC and
100kPa would be produced
from completely reacting
26.23g of magnesium with
sulfuric acid?
𝟐𝟔. 𝟕𝟓𝑳
17. How many litres of CO2
measured at 25oC and
100kPa would be produced
from completely reacting
19.36g of calcium
carbonate with nitric acid?
𝟒. 𝟕𝟗𝟓𝑳
18. How many litres of CO2
measured at 25oC and
100kPa would be produced
from completely reacting
45.66g of lithium
carbonate with
hydrochloric acid?
𝟏𝟓. 𝟑𝟐𝑳
Step 1: Write the balanced equation for the reaction.
C5H12 + 8O2 → 5CO2 + 6H2O
Step 2: Calculate the number of moles of the given substance.
𝑚𝐶5 𝐻12
235.5𝑔
𝑛𝐶5𝐻12 =
=
= 3.26421𝑚𝑜𝑙
𝑀𝐶5 𝐻12 (12.01 × 5 + 1.008 × 12)𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 3: Calculate the number of moles of the specified substance.
From the balanced equation: 𝑛𝐶𝑂2 = 5𝑛𝐶5 𝐻12 = 16.32107 𝑚𝑜𝑙
Step 4: Calculate the volume of the specified substance.
𝑣𝐶𝑂2 = 𝑛𝐶𝑂2 × 𝑉𝑀 = 16.32107𝑚𝑜𝑙 × 24.79𝐿 ⋅ 𝑚𝑜𝑙 −1 = 404.6𝐿
Step 1: Write the balanced equation for the reaction.
C8H18 + 252 O2 → 8CO2 + 9H2O
Step 2: Calculate the number of moles of the given substance.
𝑚𝐶8 𝐻18
94.78𝑔
𝑛𝐶8𝐻18 =
=
= 0.82977𝑚𝑜𝑙
𝑀𝐶8 𝐻18 (12.01 × 8 + 1.008 × 18)𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 3: Calculate the number of moles of the specified substance.
From the balanced equation: 𝑛𝐶𝑂2 = 8𝑛𝐶8 𝐻18 = 6.63818 𝑚𝑜𝑙
Step 4: Calculate the volume of the specified substance.
𝑣𝐶𝑂2 = 𝑛𝐶𝑂2 × 𝑉𝑀 = 6.63818𝑚𝑜𝑙 × 24.79𝐿 ⋅ 𝑚𝑜𝑙 −1 = 164.6𝐿
Step 1: Write the balanced equation for the reaction.
Mg + 2HCl → MgCl2 + H2
Step 2: Calculate the number of moles of the given substance.
𝑚𝑀𝑔
15.00𝑔
𝑛𝑀𝑔 =
=
= 0.61703𝑚𝑜𝑙
𝑀𝑀𝑔 24.31𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 3: Calculate the number of moles of the specified substance.
From the balanced equation: 𝑛𝐻2 = 𝑛𝑀𝑔 = 0.61703 𝑚𝑜𝑙
Step 4: Calculate the volume of the specified substance.
𝑣𝐻2 = 𝑛𝐻2 × 𝑉𝑀 = 0.61703𝑚𝑜𝑙 × 24.79𝐿 ⋅ 𝑚𝑜𝑙 −1 = 15.30𝐿
Step 1: Write the balanced equation for the reaction.
Mg + H2SO4 → MgSO4 + H2
Step 2: Calculate the number of moles of the given substance.
𝑚𝑀𝑔
26.23𝑔
𝑛𝑀𝑔 =
=
= 1.07898𝑚𝑜𝑙
𝑀𝑀𝑔 24.31𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 3: Calculate the number of moles of the specified substance.
From the balanced equation: 𝑛𝐻2 = 𝑛𝑀𝑔 = 1.07898 𝑚𝑜𝑙
Step 4: Calculate the volume of the specified substance.
𝑣𝐻2 = 𝑛𝐻2 × 𝑉𝑀 = 1.07898𝑚𝑜𝑙 × 24.79𝐿 ⋅ 𝑚𝑜𝑙 −1 = 26.75𝐿
Step 1: Write the balanced equation for the reaction.
CaCO3 + 2HNO3 → Ca(NO3)2 + H2O + CO2
Step 2: Calculate the number of moles of the given substance.
𝑚𝐶𝑎𝐶𝑂3
19.36𝑔
𝑛𝐶𝑎𝐶𝑂3 =
=
= 0.19343𝑚𝑜𝑙
𝑀𝐶𝑎𝐶𝑂3 (40.08 + 12.01 + 16.00 × 3)𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 3: Calculate the number of moles of the specified substance.
From the balanced equation: 𝑛𝐶𝑂2 = 𝑛𝐶𝑎𝐶𝑂3 = 0.19343 𝑚𝑜𝑙
Step 4: Calculate the volume of the specified substance.
𝑣𝐶𝑂2 = 𝑛𝐶𝑂2 × 𝑉𝑀 = 0.19343𝑚𝑜𝑙 × 24.79𝐿 ⋅ 𝑚𝑜𝑙 −1 = 4.795𝐿
Step 1: Write the balanced equation for the reaction.
Li2CO3 + 2HCl → 2LiCl + H2O + CO2
Step 2: Calculate the number of moles of the given substance.
𝑚𝐿𝑖2 𝐶𝑂3
45.66𝑔
𝑛𝐿𝑖2 𝐶𝑂3 =
=
= 0.61793𝑚𝑜𝑙
𝑀𝐿𝑖2 𝐶𝑂3 (6.941 × 2 + 12.01 + 16.00 × 3)𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 3: Calculate the number of moles of the specified substance.
From the balanced equation: 𝑛𝐶𝑂2 = 𝑛𝐿𝑖2 𝐶𝑂3 = 0.61793 𝑚𝑜𝑙
Step 4: Calculate the volume of the specified substance.
𝑣𝐶𝑂2 = 𝑛𝐶𝑂2 × 𝑉𝑀 = 0.61793𝑚𝑜𝑙 × 24.79𝐿 ⋅ 𝑚𝑜𝑙 −1 = 15.32𝐿
19. A hydrocarbon fuel is
completely combusted in
air to produce 15.61L of
carbon dioxide and 32.07L
of water vapour measured
at 25oC and 100kPa. What
is the empirical formula of
the hydrocarbon?
𝑪𝑯𝟒
20. A hydrocarbon fuel is
completely combusted in
air to produce 27.48L of
carbon dioxide and 13.71L
of water vapour measured
at 25oC and 100kPa. What
is the empirical formula of
the hydrocarbon?
𝑪𝑯
21. A hydrocarbon fuel is
completely combusted in
air to produce 125.32L of
carbon dioxide measured
at 25oC and 100kPa and
181.99g of water. What is
the empirical formula of
the hydrocarbon?
𝑪𝑯𝟒
22. A hydrocarbon fuel is
completely combusted in
air to produce 56.87L of
carbon dioxide measured
at 25oC and 100kPa and
31.09g of water. What is
the empirical formula of
the hydrocarbon?
𝑪𝟐 𝑯𝟑
Step 1: Calculate the number of moles of the first given substance.
𝑉𝐶𝑂2
15.61𝐿
𝑛𝐶𝑂2 =
=
= 0.62969 𝑚𝑜𝑙
𝑉𝑀
24.79𝐿 ⋅ 𝑚𝑜𝑙 −1
Step 2: Calculate the number of moles of the first element.
From the chemical formula: 𝑛𝐶 = 𝑛𝐶𝑂2 = 0.62969 𝑚𝑜𝑙
Step 3: Calculate the number of moles of the second given substance.
𝑉𝐻 𝑂
32.07𝐿
𝑛𝐻2 𝑂 = 2 =
= 1.29367 𝑚𝑜𝑙
𝑉𝑀
24.79𝐿 ⋅ 𝑚𝑜𝑙 −1
Step 4: Calculate the number of moles of the second element.
From the chemical formula: 𝑛𝐻 = 2 × 𝑛𝐻2 𝑂 = 2.58733 𝑚𝑜𝑙
Step 5: Calculate the ratio of moles first element to moles of the second element.
0.62969 2.58733
𝑛𝐶 ∶ 𝑛𝐻 = 0.62969 ∶ 2.58733 =
:
= 1: 4.10890 ≃ 1: 4
0.62969 0.62969
Step 6: Write the empirical formula. The empirical formula of the hydrocarbon is CH4
Step 1: Calculate the number of moles of the first given substance.
𝑉𝐶𝑂2
27.48𝐿
𝑛𝐶𝑂2 =
=
= 1.10851 𝑚𝑜𝑙
𝑉𝑀
24.79𝐿 ⋅ 𝑚𝑜𝑙 −1
Step 2: Calculate the number of moles of the first element.
From the chemical formula: 𝑛𝐶 = 𝑛𝐶𝑂2 = 1.10851 𝑚𝑜𝑙
Step 3: Calculate the number of moles of the second given substance.
𝑉𝐻 𝑂
13.71𝐿
𝑛𝐻2 𝑂 = 2 =
= 0.55305 𝑚𝑜𝑙
𝑉𝑀
24.79𝐿 ⋅ 𝑚𝑜𝑙 −1
Step 4: Calculate the number of moles of the second element.
From the chemical formula: 𝑛𝐻 = 2 × 𝑛𝐻2 𝑂 = 1.10609 𝑚𝑜𝑙
Step 5: Calculate the ratio of moles first element to moles of the second element.
1.10851 1.10609
𝑛𝐶 ∶ 𝑛𝐻 = 1.10851 ∶ 1.10609 =
:
= 1: 0.99782 ≃ 1: 1
1.10851 1.10851
Step 6: Write the empirical formula. The empirical formula of the hydrocarbon is CH
Step 1: Calculate the number of moles of the first given substance.
𝑉𝐶𝑂2
125.32𝐿
𝑛𝐶𝑂2 =
=
= 5.05526 𝑚𝑜𝑙
𝑉𝑀
24.79𝐿 ⋅ 𝑚𝑜𝑙 −1
Step 2: Calculate the number of moles of the first element.
From the chemical formula: 𝑛𝐶 = 𝑛𝐶𝑂2 = 5.05526 𝑚𝑜𝑙
Step 3: Calculate the number of moles of the second given substance.
𝑚𝐻2 𝑂
181.99𝑔
𝑛𝐻2 𝑂 =
=
= 10.10158 𝑚𝑜𝑙
𝑀𝐻2 𝑂 (1.008 × 2 + 16.00)𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 4: Calculate the number of moles of the second element.
From the chemical formula: 𝑛𝐻 = 2 × 𝑛𝐻2 𝑂 = 20.20315 𝑚𝑜𝑙
Step 5: Calculate the ratio of moles first element to moles of the second element.
5.05526 20.20315
𝑛𝐶 ∶ 𝑛𝐻 = 5.05526 ∶ 20.20315 =
:
= 1: 3.99646 ≃ 1: 4
5.05526 5.05526
Step 6: Write the empirical formula. The empirical formula of the hydrocarbon is CH4
Step 1: Calculate the number of moles of the first given substance.
𝑉𝐶𝑂2
56.87𝐿
𝑛𝐶𝑂2 =
=
= 2.29407 𝑚𝑜𝑙
𝑉𝑀
24.79𝐿 ⋅ 𝑚𝑜𝑙 −1
Step 2: Calculate the number of moles of the first element.
From the chemical formula: 𝑛𝐶 = 𝑛𝐶𝑂2 = 2.29407 𝑚𝑜𝑙
Step 3: Calculate the number of moles of the second given substance.
𝑚𝐻2 𝑂
31.09𝑔
𝑛𝐻2 𝑂 =
=
= 1.72569 𝑚𝑜𝑙
𝑀𝐻2 𝑂 (1.008 × 2 + 16.00)𝑔 ⋅ 𝑚𝑜𝑙 −1
Step 4: Calculate the number of moles of the second element.
From the chemical formula: 𝑛𝐻 = 2 × 𝑛𝐻2 𝑂 = 3.45138 𝑚𝑜𝑙
Step 5: Calculate the ratio of moles first element to moles of the second element.
2.29407 3.45138
𝑛𝐶 ∶ 𝑛𝐻 = 2.29407 ∶ 3.45138 =
:
= 1: 1.50448 ≃ 1: 1.5 ≃ 2: 3
2.29407 2.29407
Step 6: Write the empirical formula. The empirical formula of the hydrocarbon is C2H3