Combustion Analysis

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Combustion analysis questions commonly fall into 3 types:

1. Find C_xH_y given V_CO2, V_O2, V_CxHy
2. Find C_xH_y given ∆V, V_CO2, V_CxHy
3. Find C_xH_yO_z given m_CO2, m_H2O, m_CxHy

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1.  Find C_xH_y given V_CO2, V_O2, V_CxHy

 

C_xH_y + (x + y/4) O₂ → x CO₂ + y/2 H₂O

 

Step 1: From the question, find V_CxHy used, V_CO2 produced, V_O2,rxted.

 

Step 2: Use the ratios to solve for x and y.

 

x = VCO2 1 VCxHy

x + y/4 = VO2,rxted 1 VCxHy

 

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

 

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2.  Find C_xH_y given ∆V, V_CO2, V_CxHy

 

C_xH_y + (x + y/4) O₂ → x CO₂ + y/2 H₂O

 

∆V = V_f – V_i

 

Step 1: Determine which volumes make up V_i and V_f.

 

For most qns, it is stated that the volumes are measured at room temperatures:

V_i = V_CxHy + V_O2,rxted + _VO2,excess
V_f = V_CO2 + V_O2,excess

(V_H2O cannot be ignored if the volumes are measured above 100 °C i.e. V_f = V_f = V_CO2 + V_O2,excess + V_H2O)

 

Step 2: Equate expression to ΔV to find V_O2,reacted

 

ΔV = V_CxHy + V_O2,rxted – V_CO2

 

Step 3: Use the ratios to solve for x and y.

 

x = VCO2 1 VCxHy

x + y/4 = VO2,rxted 1 VCxHy

 

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

 

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3.  Find C_xH_y given m_CO2, m_CxHyOz, m_H2O

 

C_xH_yO_z + ? 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 m_CO2 calculate n_CO2 = n_C

m_C = (m_CO2/ 44.0) x 12.0

 

Using m_H2O calculate n_H2O = 0.5 n_H

m_H = (m_H2O/ 18.0) x 2 x 1.0

 

m_O = m_CxHyOz – m_C – m_H

 

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