Fuels Worked ExamplesIGCSE Chemistry 0620

(a) A hydrocarbon is a compound that contains hydrogen and carbon only.

(b) Check each formula for any element other than C and H. $\text{C}_2\text{H}_5\text{OH}$ contains O, $\text{CH}_3\text{Cl}$ contains Cl and $\text{CO}_2$ contains O and no H. The hydrocarbons are $\text{CH}_4$, $\text{C}_3\text{H}_6$ and $\text{C}_8\text{H}_{18}$.

(c) A mixture contains different substances that are not chemically combined. The hydrocarbons in petroleum have different boiling points, so they can be separated by boiling and condensing them at different temperatures.

(a) The fractions from the top are: refinery gas, gasoline (petrol), naphtha, kerosene (paraffin), diesel oil (gas oil), fuel oil, lubricating oil, bitumen. So J is gasoline, K is kerosene and L is lubricating oil.

(b) J: fuel for cars. K: jet fuel. L: lubricants, waxes and polishes.

(c) Naphtha is the chemical feedstock (a raw material for making other chemicals).

Naphtha is used as a chemical feedstock, so C is correct.

A and B swap two uses: bitumen is used for making roads, and fuel oil is the fuel for ships and home heating systems. D is wrong: jet fuel comes from kerosene; lubricating oil is used for lubricants, waxes and polishes.

(a) 216 °C lies in the range 180–260 °C, so it is in kerosene (12 carbon atoms also fits 10–16). $\text{C}_{12}\text{H}_{26}$ fits $\text{C}_n\text{H}_{2n+2}$ with $n = 12$ ($2 \times 12 + 2 = 26$), so it is an alkane.

(b) Going up the column (from the bottom to the top): chain length decreases, volatility increases (the fractions evaporate more easily), and viscosity decreases (they flow more easily). Boiling points get lower.

(c) Bigger molecules have stronger attractive forces between molecules, so more energy is needed to overcome them and the boiling point is higher.

Complete combustion gives carbon dioxide and water only. Balance C first, then H, then O. If you need a half $\text{O}_2$, double everything.

(a) 4 C → 4 $\text{CO}_2$; 10 H → 5 $\text{H}_2\text{O}$; O needed $= 8 + 5 = 13$ atoms $= 6\tfrac{1}{2}\;\text{O}_2$. Double:

(b) 5 $\text{CO}_2$ and 5 $\text{H}_2\text{O}$ need $10 + 5 = 15$ O atoms $= 7\tfrac{1}{2}\;\text{O}_2$. Double:

(c) 3 $\text{CO}_2$ and 4 $\text{H}_2\text{O}$ (8 H in total) contain $6 + 4 = 10$ O atoms. The alcohol already supplies 1 O, so 9 O atoms must come from oxygen $= 4\tfrac{1}{2}\;\text{O}_2$. Double:

(a) 7 C → 7 CO; 16 H → 8 $\text{H}_2\text{O}$; O atoms needed $= 7 + 8 = 15$, so $7\tfrac{1}{2}\;\text{O}_2$. Double everything:

(b) 7 C and 8 $\text{H}_2\text{O}$: only the water needs oxygen, 8 O atoms $= 4\;\text{O}_2$.

(c) There is not enough oxygen to oxidise all the carbon in the fuel to carbon dioxide, so some carbon forms carbon monoxide or carbon (soot) instead.

(d)(i) Carbon monoxide is a toxic gas. (ii) Particulates increase the risk of respiratory problems and cancer.

(a) $M_r$ of $\text{C}_4\text{H}_{10} = 4(12) + 10(1) = 58$; 2.90 kg = 2900 g.

(b) Butane : $\text{CO}_2$ = 2 : 8 = 1 : 4. $M_r$ of $\text{CO}_2 = 44$.

(c) Butane : $\text{O}_2$ = 2 : 13.

(d) Oxygen is 21% of the air, so

The converter reaction is:

A, B and C are all correct. CO gains oxygen (oxidised) and NO loses oxygen (reduced), as C says.

D is not correct: the converter turns carbon monoxide into carbon dioxide, a greenhouse gas that contributes to climate change. The answer is D.

(a) Each small division is 1 °C. Start: 21 °C. End: 47 °C.

(b) Mass of fuel 1 burned $= 152.36 - 151.46 = 0.90$ g.

(c) Fuel 2 gives the larger temperature rise per gram, so (with the same mass of water) it releases more energy per gram.

(d) Any two: use a draught shield or insulate the can to reduce heat loss; put a lid on the can; keep the same distance between flame and can; stir the water before reading the thermometer; repeat and average. (Soot on the can shows incomplete combustion, which also makes the result too low.)

(a) The boiling point increases as the number of carbon atoms increases.

(b) Heptane lies between hexane (69 °C) and octane (126 °C). Following the smooth curve through the points gives about 98 °C (any value from about 90 to 105 °C is reasonable).

(c) A substance is a gas at 25 °C if its boiling point is below 25 °C: the alkanes with 1 to 4 carbon atoms (methane, ethane, propane, butane). They are in the refinery gas fraction.

(d) Pentane distils first: it has the lowest boiling point.

(a) Divide the energy per mole by $M_r$ ($\text{CH}_4 = 16$, $\text{C}_8\text{H}_{18} = 114$, $\text{H}_2 = 2$).

(b) Methane: $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$, so 1 mol (44 g) of $\text{CO}_2$ per 890 kJ. Octane: 8 mol (352 g) of $\text{CO}_2$ per 5470 kJ.

(c) Hydrogen: it releases the most energy per gram and its only product is water — no carbon dioxide ($2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$). Disadvantage (any one): it is a gas that is difficult to store and transport; it is flammable/explosive; much hydrogen is made from fossil fuels such as methane, which releases carbon dioxide.

(a) It measures the temperature of the vapour that is passing into the condenser, which is the boiling point of the substance being collected.

(b) The alkanes are flammable; a naked flame could ignite the vapour.

(c) So the condenser fills completely with water, giving the most efficient cooling.

(d) While pentane distils, the thermometer stays at about 36 °C. When the pentane has all distilled, the temperature rises; change the tube when it reaches about 69 °C and stays steady — hexane is now distilling.

(e) Octane has longer molecules, so it is (i) more viscous and (ii) less volatile than pentane.