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A hydrocarbon with general formaul cxhy is burned completely in air yielding 0.18 g of water and 0.44 g of carbon dioxide. Which formula could give such data?

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Answer:

The empirical formula is CH2

Step-by-step explanation:

Step 1: Data given

Burning a hydrocarbon yields 0.18 grams of H2O and 0.44 grams of CO2

Molar mass of CO2 = 44 g/mol

Molar mass of H2O = 18 g/mol

Step 2: Calculate moles

Moles = Mass / molar mass

Moles CO2 = 0.44 grams / 44 g/mol = 0.01 moles

Moles H2O = 0.18 grams / 18 g/mol = 0.01 moles

In CO2 we have 1 mol C

For 0.01 mol CO2 that means we have 0.01 mol C

In H2O we have 2 mol H

For 0.01 mol H2O that means we have 2*0.01 = 0.02 mol H

Step 3: Calculate mol ratio

ratio C:H = 1:2

That means for each C atom we'll have 2 H atom

The empirical formula is CH2

User MassDebates
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6 votes

Answer:

CH2

Step-by-step explanation:

When a hydrocarbon is burnt in air or oxygen, there are two products only, these are water and carbon iv oxide.

Fuel + Oxygen—-> Water + Carbon iv oxide

We can get the formula through calculations as follows.

From the mass of carbon iv oxide produced, we can get the number of moles of carbon produced. We first divide the mass by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol

The number of moles of carbon iv oxide is 0.44/44 = 0.01

Since there is only one carbon atom in CO2, the number of moles of carbon is same as above I.e 0.01 moles

From the number of moles of water, we can get the number of moles of hydrogen. To get the number of moles of water, we need to divide the mass of water by its molar mass. Its molar mass is 18g/mol. The number of moles here is thus 0.18/18 = 0.01moles

But there are 2 atoms of hydrogen in 1 mole of water and thus, the number of moles of hydrogen is 2 * 0.01= 0.02moles

The empirical formula can be obtained by dividing the number of moles of each by the smallest which is that of 0.01

H = 0.02/0.01 = 2

C = 0.01/0.01 = 1

From the calculations, x = 1 and y = 2

The empirical formula is thus CH2

User Dyouberg
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