A compound is found to be 30.45% N and 69.55 % O by mass. If 1.63 g of this compound occupy 389 mL at 0.00 °C and 1.0332 bar, what is the molecular formula of the compound?

Question

Quizplus · Accepted Answer

1) We can first find the mass percentage of each element present in the compound: - Mass percentage of N = (30.45 g N / 100 g compound) = 0.3045 - Mass percentage of O = (69.55 g O / 100 g compound) = 0.6955 2) Next, we can calculate the number of moles of each element present in 1.63 g of the compound: - Moles of N = (0.3045)(1.63 g) / 14.01 g/mol = 0.0354 - Moles of O = (0.6955)(1.63 g) / 16.00 g/mol = 0.0718 3) We can then find the mole ratio of N to O: - Divide both mole values by the smaller value (0.0354): - N:O mole ratio ≈ 1:2 (rounded to nearest whole number) 4) We can use this ratio to write a tentative empirical formula: - Empirical formula: N1O2 5) Finally, we can use the molar mass of the empirical formula to see if we need to multiply by a whole number to get the actual molecular formula. The molar mass of the empirical formula is: - (1)(14.01 g/mol) + (2)(16.00 g/mol) = 46.01 g/mol - The actual molecular mass (found from the mass of 1.63 g and the density of the compound) is approximately 46.5 g/mol. - Since these two masses are close, we don't need to multiply the empirical formula by anything, and the molecular formula is the same as the empirical formula: - Molecular formula: N1O2₄

A Compound Is Found to Be 30