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Chapter 3: Problem 94

A compound contains only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{N}\) . Combustion of 35.0 \(\mathrm{mg}\) of the compound produces 33.5 $\mathrm{mg} \mathrm{CO}_{2}\( and 41.1 \)\mathrm{mg}\( \)\mathrm{H}_{2} \mathrm{O}$ . What is the empirical formula of the compound?

Short Answer

Expert verified
The empirical formula of the compound is CH₆N₂.

Step by step solution

01

Determine moles of C in the compound

First, we use the amount of CO₂ produced to determine the moles of C in the compound. From the balanced combustion reaction, we know that for each mole of CO₂ produced, there must be one mole of C in the original compound. The molecular weight of CO₂ is 12.01 g/mol (C) + 2 * 16.00 g/mol (O) = 44.01 g/mol. Thus, we have: Moles of CO₂ = Mass of CO₂ / Molecular weight of CO₂ Moles of CO₂ = 33.5 mg / 44.01 g/mol Since 1g = 1000mg: Moles of CO₂ = 33.5 * 10^(-3) g / 44.01 g/mol = 0.00076 mol Since the moles of CO₂ are equal to the moles of C, the moles of C in the compound is 0.00076 mol.
02

Determine moles of H in the compound

Next, we use the amount of H₂O produced to determine the moles of H in the compound. From the balanced combustion reaction, we know that for each mole of H₂O produced, there must be two moles of H in the original compound. The molecular weight of H₂O is 2 * 1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol. Thus, we have: Moles of H₂O = Mass of H₂O / Molecular weight of H₂O Moles of H₂O = 41.1 mg / 18.02 g/mol Since 1g = 1000mg: Moles of H₂O = 41.1 * 10^(-3) g / 18.02 g/mol = 0.00228 mol And because 1 mole of H₂O contains 2 moles of H, Moles of H = 2 * Moles of H₂O = 2 * 0.00228 mol = 0.00456 mol
03

Determine mass and moles of N in the compound

To determine the mass of N in the compound, we'll use the initial mass of the compound and subtract the masses of C and H. Mass of N = Mass of compound - (Mass of C + Mass of H) Mass of C = Moles of C * Molecular weight of C = 0.00076 mol * 12.01 g/mol = 9.13 * 10^(-3) g Mass of H = Moles of H * Molecular weight of H = 0.00456 mol * 1.01 g/mol = 4.60 * 10^(-3) g Since the given mass of compound is 35.0mg or 0.035g, Mass of N = 0.035 g - (9.13 * 10^(-3) g + 4.60 * 10^(-3) g) = 0.02127 g Now, we can find the moles of N using its molecular weight (14.01 g/mol): Moles of N = Mass of N / Molecular weight of N = 0.02127 g / 14.01 g/mol = 0.00152 mol
04

Determine the empirical formula

Now that we have determined the moles of C, H, and N in the compound, we can find the simplest whole number ratio of these elements. First, we divide each element's moles by the smallest number of moles to get a preliminary ratio: Preliminary ratio of C : H : N = 0.00076 : 0.00456 : 0.00152 Divide by the smallest moles (C here): = 1 : 6 : 2 Thus, the empirical formula of the compound is CH₆N₂.

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Most popular questions from this chapter

A sample of a hydrocarbon (a compound consisting of only carbon and hydrogen contains \(2.59 \times 10^{23}\) atoms of hydrogen and is 17.3\(\%\) hydrogen by mass. If the molar mass of the hydrocarbon is between 55 and 65 g/mol, what amount (moles) of compound is present, and what is the mass of the sample?

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