Question

A sample of hydrocarbon undergoes complete combustion to produce carbon dioxide (CO2) and
water (H2O).
(Note: Atomic mass of C = 12 amu, O = 16 amu and H = 1 amu)
If 8.80 g of carbon dioxide is obtained, then what will be the mass of carbon present in it?

Answer 1

Answer:

example

Explanation:

1.5 g of hydrocarbon undergoes complete combustion to give 4.4 g of CO2 and 2.7 g of H2O. Given this data, what is the empirical formula of this hydrocarbon.

The first step here is to determine the mass of C in CO2 and the mass of H in H2O. This is done by dividing the atomic mass by the molecular mass and then multiplying by the mass of compound produced.

For C:

(12.011 g / 44.009 g) x 4.40 g = 1.1999 g

For H:

(1.0079 x 2 / 18.0148 g) x 2.70 g = 0.3021 g

The next step is to convert these masses into moles. This is done by dividing the mass by the relative atomic mass of the element:

C: 1.1999 g / 12.011 g mol-1 = 0.0999 mol

H: 0.3021 g / 1.0079 g mol-1 = 0.2997 mol

The final step is to divide each of these two values by the smallest number, in this case this is the number of moles of carbon:

C: 0.0999 mol / 0.0999 mol = 1

H: 0.2997 mol / 0.0999 mol = 3

We therefore have a ratio of 1 carbon atom to 3 hydrogen atoms, thus the empirical formula for this hydrocarbon is CH3.

Answer 2

Answer:

Carbondioxide (CO2) -4. 4g

water(H2O) - 2.7g

Explanation:

1.5 g of hydrocarbon undergoes complete combustion.

As the embirical formula for hydrocarbon is C7H8.

It reacts with oxygen. Hence,carbondioxide and water are released.

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