Question

A 42.0g sample of compound containing only C and H was analyzed. The results showed that the sample contained 36.0g of C and 6.0g of H. Which of the following questions about the compound can be answered using the results of the analysis?

Answer 1

Answer:

We can solve the question 'What is the empirical formula of the compound?' The answer is CH2

Explanation:

A 42.0g sample of compound containing only C and H was analyzed. The results showed that the sample contained 36.0g of C and 6.0g of H. Which of the following questions about the compound can be answered using the results of the analysis?

A) What was the volume of the sample?

B) What is the molar mass of the compound?

C) What is the chemical stability of the compound?

D) What is the empirical formula of the compound?

Step 1: Data given

Mass of the compound = 42.0 grams

⇒ 36.0 grams = carbon

⇒ 6.0 grams = hydrogen

Molar mass of carbon = 12.01 g/mol

Molar mass of hydrogen = 1.01 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles Carbon = 36.0 grams / 12.01 g/mol

Moles carbon = 3.00 moles

Moles hydrogen = 6.0 grams / 1.01 g/mol

Moles hydrogen = 5.95 moles

Step 3: Calculate mol ratio

We divide by the smallest number of moles

Carbon: 3.00 / 3.00 = 1

Hydrogen: 5.95 / 3.00 = 2

The empirical formula is CH2

Answer 2

Molar mass of C_3H_6 = 36 g mol^-1 +6g mol^-1 =42g mol^-1

Empirical formula of Sample = (CH_3)

Explanation:

Mass sample having only C and H = 42g

Mass of C = 12g mol^-1

Molar mass of C = 12g mol^-1

Moles of C = 36 g / 12g mol^-1

= 3 mol

Mass of H in sample = 6g

Molar mass of H = 1 g mol^-1

Moles of H = 6g / 1g mol^-1

= 6 mol

Therefore in 42g sample 3 mol of C + 6 mol of H present, that is C_3H_6

Molar mass of C_3H_6 = 36 g mol^-1 +6g mol^-1 =42g mol^-1

Empirical formula of Sample = (CH_3)