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Chapter 5: Problem 92

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 \(\mathrm{CH}_6 \mathrm{N}_2\).

Step by step solution

01

Calculate the amount of carbon in the compound

To find the amount of carbon in the compound, first, let's determine the amount of carbon present in \(\mathrm{CO}_2\). Since the molar mass of carbon is 12.01g/mol and the molar mass of \(\mathrm{CO}_2\) is 44.01 g/mol, we get the proportion of carbon in \(\mathrm{CO}_2\) as: \[ \frac{12.01}{44.01} \approx 0.273 \] Now, multiply the mass of \(\mathrm{CO}_2\) produced (33.5 mg) by the proportion of carbon in \(\mathrm{CO}_2\) to find the mass of carbon in the compound: \[ 33.5 \mathrm{mg} \cdot 0.273 \approx 9.14 \mathrm{mg} \]
02

Calculate the amount of hydrogen in the compound

Similarly, we'll find the amount of hydrogen in the compound. Since the molar mass of hydrogen is 1.01 g/mol and the molar mass of \(\mathrm{H}_2 \mathrm{O}\) is 18.02 g/mol, we get the proportion of hydrogen in \(\mathrm{H}_2 \mathrm{O}\) as: \[ \frac{2 \cdot 1.01}{18.02} \approx 0.112 \] Now, multiply the mass of \(\mathrm{H}_2 \mathrm{O}\) produced (41.1 mg) by the proportion of hydrogen in \(\mathrm{H}_2 \mathrm{O}\) to find the mass of hydrogen in the compound: \[ 41.1 \mathrm{mg} \cdot 0.112 \approx 4.60 \mathrm{mg} \]
03

Calculate the amount of nitrogen in the compound

The compound contains only C, H, and N. We know the total mass of the compound is 35.0 mg. So, to find the mass of nitrogen, we subtract the masses of carbon and hydrogen we calculated in steps 1 and 2: \[ 35.0 \mathrm{mg} - 9.14 \mathrm{mg} - 4.60 \mathrm{mg} \approx 21.26 \mathrm{mg} \]
04

Find moles of each element

To determine the empirical formula, we need to find the simplest whole number ratio of each element in the compound. For this, divide the mass of each element by its respective molar mass to find the moles: Moles of carbon: \[ \frac{9.14 \mathrm{mg}}{12.01 \mathrm{g/mol} \cdot 1000 \mathrm{mg/g}} \approx 0.760 \mathrm{mol} \] Moles of hydrogen: \[ \frac{4.60 \mathrm{mg}}{1.01 \mathrm{g/mol} \cdot 1000 \mathrm{mg/g}} \approx 4.56 \mathrm{mol} \] Moles of nitrogen: \[ \frac{21.26 \mathrm{mg}}{14.01 \mathrm{g/mol} \cdot 1000 \mathrm{mg/g}} \approx 1.52 \mathrm{mol} \]
05

Find the simplest whole number ratio of elements

Divide the moles of each element by the smallest number of moles (0.760 in this case) to find the simplest whole number ratio: \[ \frac{0.760}{0.760} : \frac{4.56}{0.760} : \frac{1.52}{0.760} \approx 1 : 6 : 2 \]
06

Write the empirical formula

Based on the simplest whole number ratio of elements, the empirical formula of the compound is: \[ \mathrm{CH}_6 \mathrm{N}_2 \]

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Most popular questions from this chapter

Phosphorus can be prepared from calcium phosphate by the following reaction: $$\begin{aligned}2 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6 \mathrm{SiO}_{2}(s)+& 10 \mathrm{C}(s) \longrightarrow \\\& 6 \mathrm{CaSiO}_{3}(s)+\mathrm{P}_{4}(s)+10 \mathrm{CO}(g) \end{aligned}$$ Phosphorite is a mineral that contains \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) plus other non-phosphorus- containing compounds. What is the maximum amount of \(\mathrm{P}_{4}\) that can be produced from \(1.0 \mathrm{kg}\) of phosphorite if the phorphorite sample is \(75 \% \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) by mass? Assume an excess of the other reactants.

A common demonstration in chemistry courses involves adding a tiny speck of manganese(IV) oxide to a concentrated hydrogen peroxide \(\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)\) solution. Hydrogen peroxide decomposes quite spectacularly under these conditions to produce oxygen gas and steam (water vapor). Manganese(IV) oxide is a catalyst for the decomposition of hydrogen peroxide and is not consumed in the reaction. Write the balanced equation for the decomposition reaction of hydrogen peroxide.

One of the components that make up common table sugar is fructose, a compound that contains only carbon, hydrogen, and oxygen. Complete combustion of \(1.50 \mathrm{g}\) of fructose produced \(2.20 \mathrm{g}\) of carbon dioxide and \(0.900 \mathrm{g}\) of water. What is the empirical formula of fructose?

A 0.755-g sample of hydrated copper(II) sulfate $$\mathrm{CuSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O}$$ was heated carefully until it had changed completely to anhydrous copper(II) sulfate (CuSO_) with a mass of 0.483 g. Determine the value of \(x .\) [This number is called the number of waters of hydration of copper(II) sulfate. It specifies the number of water molecules per formula unit of \(\mathrm{CuSO}_{4}\) in the hydrated crystal.]

The compound \(\mathrm{As}_{2} \mathrm{I}_{4}\) is synthesized by reaction of arsenic metal with arsenic triiodide. If a solid cubic block of arsenic \(\left(d=5.72 \mathrm{g} / \mathrm{cm}^{3}\right)\) that is \(3.00 \mathrm{cm}\) on edge is allowed to react with \(1.01 \times 10^{24}\) molecules of arsenic triiodide, what mass of \(\mathrm{As}_{2} \mathrm{I}_{4}\) can be prepared? If the percent yield of \(\mathrm{As}_{2} \mathrm{I}_{4}\) was \(75.6 \%\) what mass of \(\mathrm{As}_{2} \mathrm{I}_{4}\) was actually isolated?
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