Question

1.50 g sample of hydrocarbon undergoes complete combustion to produce

4.40 g of CO2 and 2.70 g of H2O. What is the empirical formula of this compound? In

addition, its molecular weight has been determined to be about 78. What is the

molecular formula?​

Answer 1

Answer:

: A 1.50 g sample of hydrocarbon undergoes complete combustion to produce 4.40 g of CO2 and

2.70 g of H2O. What is the empirical formula of this compound?

Solution:

1) Determine the grams of carbon in 4.40 g CO2 and the grams of hydrogen in 2.70 g H2O.

carbon: 4.40 g x (12.011 g / 44.0098 g) = 1.20083 g

hydrogen: 2.70 g x (2.0158 g / 18.0152 g) = 0.3021482 g

2) Convert grams of C and H to their respective amount of moles.

carbon: 1.20083 g / 12.011 g/mol = 0.09998 mol

hydrogen: 0.3021482 g / 1.0079 g/mol = 0.2998 mol

3) Divide each molar amount by the lowest value, seeking to modify the above molar amounts into small, whole

numbers.

carbon: 0.09998 mol / 0.09998 mol = 1

hydrogen: 0.2998 mol / 0.09998 mol = 2.9986 = 3

We have now arrived at the answer: the empirical formula of the substance is CH3

Explanation:

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