Qeustion 1: A compound containing only C, H and O was subjected to combustion an

Question

Qeustion 1: A compound containing only C, H and O was subjected to combustion analysis. A sample of 6.410×10-2g produced 6.266×10-2g of CO2and 1.283×10-2g of H2O. Determine the empirical formula of the compound and enter the appropriate subscript after each element.

C(subscript) H(subscript) O(subscript)

If the molar mass of the compound is 45.017 g/mol, determine the molecular formula of the compound and enter the appropriate subscript after each element.

the answer is? C(subscript) H(subscript) O(subscript)

Explanation / Answer

Qeustion 1: A compound containing only C, H and O was subjected to combustion analysis. A sample of 6.410×10-2 g produced 6.266×10-2 g of CO2 and 1.283×10-2 g of H2O. Determine the empirical formula of the compound and enter the appropriate subscript after each element.

Solution :-

Lets calculate the moles of the CO2 and H2O

Moles of CO2 = 0.06266 g / 44.01 g per mol = 0.001424 mol

Moles of H2O = 0.01283 g / 18.015 g per mol = 0.000712 mol

Now lets calculate the moles of the C and H

0.001424 mol CO2 * 1 mol C / 1 mol CO2 = 0.001424 mol C

0.000712 mol H2O * 2 mol H / 1 mol H2O = 0.001424 mol H

Now lets calculate the mass of the C and H

Mass of C= 0.001424 mol * 12.01 g per mol = 0.0171 g C

Mass of H = 0.001424 mol * 1.0079 g per mol = 0.001435 g

Now lets calculate the mass of oxygen

Mass of oxygen = 0.06410 g – ( mass of C+ mass of H)

                           = 0.06410 g – (0.0171 g +0.001435 g)

                           = 0.045565 g O

Now lets find moles of the oxygen

Moles of O = 0.045565 g / 16 g per mol = 0.002848 mol

Now lets find the mole ratio of the elements

C= 0.001424 / 0.001424 = 1

H= 0.001424/0.001424 = 1

O = 0.002848 /0.001424 = 2

So the empirical formula is = CHO2

If the molar mass of the compound is 45.017 g/mol, determine the molecular formula of the compound and enter the appropriate subscript after each element.

The answer is? C(subscript) H(subscript) O(subscript)

Solution :-

Lets find the ratio of the molar mass to empirical formula mass

Empirical formula mass for the CHO2 = 45.02 g per mol

Molar mass / empirical formula mass = 45.017 g per mol / 45.02 g per mol = 1

So the molecular formula = CHO2

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