What is the empirical formula of a compound containing 90 grams carbon, 11 grams hydrogen, and 35 grams nitrogen? a. c3h4n b. c2h4n c. c4h10n d. c2h5n

A. C₃H₄N

Explanation:

Firstly, we need to calculate the no. of moles of C, H, and N using the relation:

no. of moles = mass/molar mass.

∴ no. of moles of C = mass/molar mass = (90.0 g)/(12.0 g/mol) = 7.5 mol.

∴ no. of moles of H = mass/molar mass = (11.0 g)/(1.0 g/mol) = 11.0 mol.

∴ no. of moles of N = mass/molar mass = (35.0 g)/(14.0 g/mol) = 2.5 mol.

We should get the mole ratio of each atom by dividing by the lowest no. of moles (2.5 mol of N).

∴ the mole ratio of C: H: N = (7.5 mol/2.5 mol): (11.0 mol/2.5 mol): (2.5 mol/2.5 mol) = (3: 4.4: 1) ≅ (3: 4: 1).

So, the empirical formula is: A. C₃H₄N.

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