Question

A 1.0 g sample of hydrogen reacts completely with 19.0 g of fluorine to form a compound of hydrogen and fluorine. a. What is the percent by mass of each element in the compound? b. What mass of hydrogen would be present in a 50 g sample of this compound? c. Justify your answer to b.

Answer 1

Explanation:tr

a) Molar mass of HF = 20 g/mol

Atomic mass of hydrogen = 1 g/mol

Atomic mass of fluorine = 19 g/mol

Percentage of an element in a compound:

$\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{Molar mass of compound }}\times 100$

Percentage of fluorine:

$\frac{1\times 19 g/mol}{20g/mol}\times 100=95\%$

Percentage of hydrogen:

$\frac{1\times 1g/mol}{20 g/mol}\times 100=5\%$

b) Mass of hydrogen in 50 grams of HF sample.

Moles of HF = $\frac{50 g}{20 g/mol}=2.5 mol$

1 mole of HF has 1 mole of hydrogen atom.

Then 2.5 moles of HF will have:

$1\times 2.5 mol=2.5 mol$ of hydrogen atom.

Mass of 2.5 moles of hydrogen atom:

1 g/mol × 2.5 mol = 2.5 g

2.5 grams of hydrogen would be present in a 50 g sample of this compound.

c) As we solved in part (a) that HF molecules has 5% of hydrogen by mass.

Then mass of hydrogen in 50 grams of HF compound we will have :

5% of 50 grams of HF = $\frac{5}{100}\times 50 g=2.5 g$