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Chapter 3: Problem 46

Determine the empirical formula of each of the following compounds if a sample contains (a) \(3.92 \mathrm{~mol} \mathrm{C}, 5.99 \mathrm{~mol} \mathrm{H},\) and \(2.94 \mathrm{~mol} \mathrm{O} ;\) (b) \(12.0 \mathrm{~g}\) calcium and 2.8 g nitrogen; \((\mathbf{c})\) \(89.14 \%\) Au and \(10.86 \%\) O by mass.

Short Answer

Expert verified
(a) For given moles, we find the ratios to be approximately 4:6:3. Hence, the empirical formula is \(C_4H_6O_3\). (b) For given masses, we find the ratios to be approximately 1:1. Therefore, the empirical formula is \(CaN\). (c) For given percentages, we find the ratios to be approximately 2:3. Hence, the empirical formula is \(Au_2O_3\).

Step by step solution

01

a) Calculate moles of elements and their ratios

Given the moles of elements are: - Carbon (C): 3.92 mol - Hydrogen (H): 5.99 mol - Oxygen (O): 2.94 mol Calculate the mole ratios by dividing each value by the smallest number of moles (2.94). C: \(\frac{3.92}{2.94} = 1.33\) (approx.) H: \(\frac{5.99}{2.94} = 2.04\) (approx.) O: \(\frac{2.94}{2.94} = 1\)
02

a) Determine the empirical formula

The ratios are close to 4:6:3, after multiplying each ratio by 3. Therefore, the empirical formula is \(C_4H_6O_3\).
03

b) Convert masses to moles and calculate ratios

Given the masses of elements are: - Calcium (Ca): 12.0 g - Nitrogen (N): 2.8 g First, we need to convert the masses to moles using the molar mass of each element: (Ca: 40.08 g/mol, N: 14.01 g/mol) Ca: \(\frac{12.0}{40.08}= 0.299\) mol (approx.) N: \(\frac{2.8}{14.01}= 0.200\) mol (approx.) Calculate the mole ratios by dividing each value by the smallest number of moles (0.200). Ca: \(\frac{0.299}{0.200}=1.495\) (approx.) N: \(\frac{0.200}{0.200}=1\)
04

b) Determine the empirical formula

The ratios are close to 1:1, so the empirical formula is \(CaN\).
05

c) Convert percentages to masses and moles

To solve this problem, we can assume having a 100 g sample, where the mass of each element is: - Gold (Au): 89.14 g (89.14% of 100 g) - Oxygen (O): 10.86 g (10.86% of 100 g) Now, convert the masses to moles using the molar mass of each element: (Au: 197.0 g/mol, O: 16.00 g/mol) Au: \(\frac{89.14}{197.0} = 0.452\) mol (approx.) O: \(\frac{10.86}{16.00} = 0.679\) mol (approx.) Calculate the mole ratios by dividing each value by the smallest number of moles (0.452). Au: \(\frac{0.452}{0.452} = 1\) O: \(\frac{0.679}{0.452} = 1.50\) (approx.)
06

c) Determine the empirical formula

The ratios are close to 2:3 after multiplying by 2. Therefore, the empirical formula is \(Au_2O_3\).

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Most popular questions from this chapter

Balance the following equations: (a) $\mathrm{CaS}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{HS})_{2}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q)$ (b) $\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)$ (c) $\mathrm{FeCl}_{3}(s)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow \mathrm{Fe}_{2}\left(\mathrm{CO}_{3}\right)_{3}(s)+\mathrm{NaCl}(a q)$ (d) $\mathrm{FeS}_{2}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{2}(g)$

Calcium hydride reacts with water to form calcium hydroxide and hydrogen gas. (a) Write a balanced chemical equation for the reaction. (b) How many grams of calcium hydride are needed to form \(4.500 \mathrm{~g}\) of hydrogen?

(a) Combustion analysis of toluene, a common organic solvent, gives $5.86 \mathrm{mg}\( of \)\mathrm{CO}_{2}\( and \)1.37 \mathrm{mg}\( of \)\mathrm{H}_{2} \mathrm{O} .$ If the compound contains only carbon and hydrogen, what is its empirical formula? (b) Menthol, the substance we can smell in mentholated cough drops, is composed of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\). A \(0.1005-g\) sample of menthol is combusted, producing \(0.2829 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.1159 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula for menthol? If menthol has a molar mass of $156 \mathrm{~g} / \mathrm{mol}$, what is its molecular formula?

Write balanced chemical equations to correspond to each of the following descriptions: (a) When sulfur trioxide gas reacts with water, a solution of sulfuric acid forms. (b) Boron sulfide, \(\mathrm{B}_{2} \mathrm{~S}_{3}(s),\) reacts violently with water to form dissolved boric acid, $\mathrm{H}_{3} \mathrm{BO}_{3},$ and hydrogen sulfide gas. (c) Phosphine, \(\mathrm{PH}_{3}(g)\), combusts in oxygen gas to form water vapor and solid tetraphosphorus decaoxide. (d) When solid mercury(II) nitrate is heated, it decomposes to form solid mercury(II) oxide, gaseous nitrogen dioxide, and oxygen. (e) Copper metal reacts with hot concentrated sulfuric acid solution to form aqueous copper(II) sulfate, sulfur dioxide gas, and water.

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(74.0 \% \mathrm{C}, 8.7 \% \mathrm{H},\) and \(17.3 \% \mathrm{~N}\) (b) \(57.5 \% \mathrm{Na}, 40.0 \% \mathrm{O},\) and \(2.5 \% \mathrm{H}\) (c) \(41.1 \% \mathrm{~N}, 11.8 \% \mathrm{H},\) and the remainder \(\mathrm{S}\)
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