Combustion Analysis
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Combustion analysis questions commonly fall into 3 types:
- Find CxHy given VCO2, VO2, VCxHy
- Find CxHy given ∆V, VCO2, VCxHy
- Find CxHyOz given mCO2, mH2O, mCxHy
1. Find CxHy given VCO2, VO2, VCxHy
CxHy + (x + y/4) O2 → x CO2 + y/2 H2O
Step 1: From the question, find VCxHy used, VCO2 produced, VO2,rxted.
Step 2: Use the ratios to solve for x and y.
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| Example 1: 10 cm3 of a gaseous hydrocarbon underwent complete combustion in 80 cm3 of oxygen, the remaining gases occupied 60 cm3. On passing the gases through NaOH, the volume decreased by 40 cm3. Determine the molecular formula of the hydrocarbon (All volumes were measured under room conditions).
VCxHy = 10 cm3 VCO2 = 40 cm3 (CO2 reacts with NaOH) VO2,excess = 60 – 40 = 20 cm3 VO2,reacted = 80 – 20 = 60 cm3
x = 40/10 = 4 x + y/4 = 60/10 = 6 → y = 8
Molecular formula = C4H8 |
2. Find CxHy given ∆V, VCO2, VCxHy
CxHy + (x + y/4) O2 → x CO2 + y/2 H2O
∆V = Vf – Vi
Step 1: Determine which volumes make up Vi and Vf.
For most qns, it is stated that the volumes are measured at room temperatures:
Vi = VCxHy + VO2,rxted + VO2,excess
Vf = VCO2 + VO2,excess
(VH2O cannot be ignored if the volumes are measured above 100 °C i.e. Vf = Vf = VCO2 + VO2,excess + VH2O)
Step 2: Equate expression to ΔV to find VO2,reacted
ΔV = VCxHy + VO2,rxted – VCO2
Step 3: Use the ratios to solve for x and y.
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| Example 2: When 10 cm3 of a gaseous hydrocarbon underwent combustion in excess oxygen, there was a contraction of 30 cm3. A further contraction of 40 cm3 occurred when the gases were passed through NaOH. Determine the molecular formula of the hydrocarbon (All volumes were measured under room conditions).
ΔV = VCxHy + VO2,rxted – VCO2
30 = 10 + VO2,rxted – 40
VO2,reacted = 60 cm3
Using method in Example 1, Molecular formula = C4H8 |
3. Find CxHy given mCO2, mCxHyOz, mH2O
CxHyOz + ? O2 → x CO2 + y/2 H2O
Step 1: From the question, find
| mCO2 produced | → absorbed by NaOH/ KOH |
| mH2O produced | → absorbed by P4O10 |
| mCxHy used |
Step 2: Find the masses of C, H and O.
Using mCO2 calculate nCO2 = nC
mC = (mCO2/ 44.0) x 12.0
Using mH2O calculate nH2O = 0.5 nH
mH = (mH2O/ 18.0) x 2 x 1.0
mO = mCxHyOz – mC – mH
Step 3: Use the masses to find the moles of C, H and O hence, the simplest ratio.
| C | H | O | |
| m | |||
| n = (m/Mr) | |||
| ratio |
| Example 3: When 0.160 g of CxHyOz underwent complete combustion, 0.365 g of CO2 and 0.149 g of H2O was obtained. Given that its molar mass is 116 g mol–1, determine the molecular formula of the compound.
mC = (mCO2/ 44.0) x 12.0 = (0.365/ 44.0) x 12.0 = 0.09954 g
mH = (mH2O/ 18.0) x 2 x 1.0 = (0.149/ 18.0) x 2 x 1.0 = 0.01655 g
mO = mCxHyOz – mC – mH = 0.160 – 0.09954 – 0.01655 = 0.04391 g |
| C | H | O | |
| m/ g | 0.09954 | 0.01655 | 0.04391 |
| n/ mol | 0.008295 | 0.01655 | 0.002744 |
| ratio | 3 | 6 | 1 |
| Empirical formula = C3H6O
Molar mass = n (12x3 + 6 + 16) = 116 → n = 2
Molecular formula = C6H12O2 |
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