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Chapter 3: Problem 73

Express the composition of each of the following compounds as the mass percents of its elements. a. formaldehyde, \(\mathrm{CH}_{2} \mathrm{O}\) b. glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) c. acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

Short Answer

Expert verified
a. Formaldehyde, \(\mathrm{CH}_{2} \mathrm{O}\): - Carbon: \(40.0\%\) - Hydrogen: \(6.7\%\) - Oxygen: \(53.3\%\) b. Glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\): - Carbon: \(40.0\%\) - Hydrogen: \(6.7\%\) - Oxygen: \(53.3\%\) c. Acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\): - Hydrogen: \(4.1\%\) - Carbon: \(39.9\%\) - Oxygen: \(56.0\%\)

Step by step solution

01

Calculate the molar mass of the compounds

Calculate the molar mass of each compound by adding the molar masses of their individual constituent elements. For example, the molar mass of formaldehyde (\(\mathrm{CH}_{2} \mathrm{O}\)): - Carbon: \(1 \times 12.01 \, \mathrm{g/mol} = 12.01 \, \mathrm{g/mol}\) - Hydrogen: \(2 \times 1.01 \, \mathrm{g/mol} = 2.02 \, \mathrm{g/mol}\) - Oxygen: \(1 \times 16.00 \, \mathrm{g/mol} = 16.00 \, \mathrm{g/mol}\) Apply the same calculation for glucose (\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)) and acetic acid (\(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)).
02

Calculate the mass percent of each element

To find the mass percent of each element, divide the total mass of the element in the compound by the total molar mass of the compound, then multiply by 100. For example, the mass percent of the elements in formaldehyde: - Carbon: \(\dfrac{12.01 \, \mathrm{g/mol}}{12.01+2.02+16.00 \, \mathrm{g/mol}} \times 100\% = 40.0\%\) - Hydrogen: \(\dfrac{2.02 \, \mathrm{g/mol}}{12.01+2.02+16.00 \, \mathrm{g/mol}} \times 100\% = 6.7\%\) - Oxygen: \(\dfrac{16.00 \, \mathrm{g/mol}}{12.01+2.02+16.00 \, \mathrm{g/mol}} \times 100\% = 53.3\%\) Apply the same calculation for each element in glucose (\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)) and acetic acid (\(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)).
03

Final Answer

a. Formaldehyde, \(\mathrm{CH}_{2} \mathrm{O}\): - Carbon: \(40.0\%\) - Hydrogen: \(6.7\%\) - Oxygen: \(53.3\%\) b. Glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\): - Calculate the mass percent for carbon, hydrogen, and oxygen in glucose following steps 1 and 2. c. Acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\): - Calculate the mass percent for hydrogen, carbon, and oxygen in acetic acid following steps 1 and 2.

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Most popular questions from this chapter

A sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) contains \(2.59 \times 10^{2.3}\) atoms of hydrogen and is \(17.3 \%\) hydrogen by mass. If the molar mass of the hydrocarbon is between 55 and \(65 \mathrm{~g} / \mathrm{mol}\), what amount (moles) of compound is present, and what is the mass of the sample?

A \(0.755-\mathrm{g}\) 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 \(\left(\mathrm{CuSO}_{4}\right)\) with a mass of \(0.483 \mathrm{~g}\). 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.]

A sample of urea contains \(1.121 \mathrm{~g} \mathrm{~N}, 0.161 \mathrm{~g} \mathrm{H}, 0.480 \mathrm{~g} \mathrm{C}\), and \(0.640 \mathrm{~g} \mathrm{O} .\) What is the empirical formula of urea?

Consider the reaction $$ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) $$ Identify the limiting reagent in each of the reaction mixtures given below: a. 50 molecules of \(\mathrm{H}_{2}\) and 25 molecules of \(\mathrm{O}_{2}\) b. 100 molecules of \(\mathrm{H}_{2}\) and 40 molecules of \(\mathrm{O}_{2}\) c. 100 molecules of \(\mathrm{H}_{2}\) and 100 molecules of \(\mathrm{O}_{2}\) d. \(0.50 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.75 \mathrm{~mol} \mathrm{O}\). e. \(0.80 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.75 \mathrm{~mol} \mathrm{O}_{2}\) f. \(1.0 \mathrm{~g} \mathrm{H}_{2}\) and \(0.25 \mathrm{~mol} \mathrm{O}_{2}\) g. \(5.00 \mathrm{~g} \mathrm{H}_{2}\) and \(56.00 \mathrm{~g} \mathrm{O}_{2}\)

Many cereals are made with high moisture content so that the cereal can be formed into various shapes before it is dried. A cereal product containing \(58 \% \mathrm{H}_{2} \mathrm{O}\) by mass is produced at the rate of \(1000 . \mathrm{kg} / \mathrm{h} .\) What mass of water must be evaporated per hour if the final product contains only \(20 . \%\) water?
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