Question

A 0.67 mg sample of nitrogen reacts with hydrogen to form 0.8146 mg of hydride. What is the empirical formula of the nitrogen hydride?

Answer 1

first we have to calculate the amount of hydrogen reacted = (mass of hydride) - (mas of nitrogen) = (0.8146 - 0.67)mg = 0.1446 mg

***here we have considered that all of the nitrogen is converted to product.

now we have to convert the mass of nitrogen and hydrogen into number of moles

0.67mg = 0.67 *10^-3 g of nitrogen

molecular weight of nitrogen = 14 g/mol

hence 0.67 *10^-3 g of nitrogen = (0.67 *10^-3) / 14 = 4.79 * 10^-5 mol

and atomic weight of hydrogen is 1, thus 0.1446 mg = 0.1446 * 10^-3 mol

so 4.79 * 10^-5 mol of nitrogen reacts with 0.1446 * 10^-3 mol of hydrogen

1 mol of nitrogen reacts with = (0.1446 * 10^-3 ) / (4.79 * 10^-5) = 3 moles of hydrogen

thus the reaction is: N + 3H ----> NH₃

the empirical formula of the nitrogen hydride is NH₃