Question

Calculate the percentage by mass of the indicated element in the following compounds: (a) carbon in acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\), a gas used in welding; (b) hydrogen in ascorbic acid, \(\mathrm{HC}_{6} \mathrm{H}_{7} \mathrm{O}_{6}\), also known as vitamin \(\mathrm{C} ;\) (c) hydrogen in ammonium sulfate, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\), a substance used as a nitrogen fertilizer; (d) platinum in \(\mathrm{Pt} \mathrm{Cl}_{2}\left(\mathrm{NH}_{3}\right)_{2}\), a chemotherapy agent called cisplatin; (e) oxygen in the female sex hormone estradiol, \(\mathrm{C}_{18} \mathrm{H}_{24} \mathrm{O}_{2} ;\) (f) carbon in capsaicin, \(\mathrm{C}_{18} \mathrm{H}_{27} \mathrm{NO}_{3}\), the com- pound that gives the hot taste to chili peppers.

Short answer

The percentage by mass of the indicated element in each compound is as follows: (a) Carbon in acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\): 92.25%; (b) Hydrogen in ascorbic acid, \(\mathrm{HC}_{6}\mathrm{H}_{7}\mathrm{O}_{6}\): 4.01%; (c) Hydrogen in ammonium sulfate, \(\left(\mathrm{NH}_{4}\right)_{2}\mathrm{SO}_{4}\): 6.11%; (d) Platinum in \(\mathrm{Pt} \mathrm{Cl}_{2}\left(\mathrm{NH}_{3}\right)_{2}\), cisplatin: 65.01%; (e) Oxygen in estradiol, \(\mathrm{C}_{18} \mathrm{H}_{24} \mathrm{O}_{2}\): 11.75%; (f) Carbon in capsaicin, \(\mathrm{C}_{18}\mathrm{H}_{27}\mathrm{NO}_{3}\): 70.74%.

Step-by-step solution

(a) Carbon in acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\)

1. Calculate the molar mass of acetylene: \(\mathrm{C}_{2} \mathrm{H}_{2}\) has 2 Carbon atoms and 2 Hydrogen atoms. Molar mass = (2 × 12.01 \(\mathrm{g/mol}\)) + (2 × 1.01 \(\mathrm{g/mol}\)) = 26.04 \(\mathrm{g/mol}\) 2. Calculate the mass of carbon in the compound. Mass of carbon = 2 × 12.01 \(\mathrm{g/mol}\) = 24.02 \(\mathrm{g/mol}\) 3. Calculate the percentage by mass of carbon. Percentage by mass = (24.02 / 26.04) × 100 = 92.25%

(b) Hydrogen in ascorbic acid, \(\mathrm{HC}_{6}\mathrm{H}_{7}\mathrm{O}_{6}\)

1. Calculate the molar mass of ascorbic acid: \(\mathrm{HC}_{6}\mathrm{H}_{7}\mathrm{O}_{6}\) has 1 Carbon atom, 7 Hydrogen atoms, and 6 Oxygen atoms. Molar mass = (1 × 12.01 \(\mathrm{g/mol}\)) + (7 × 1.01 \(\mathrm{g/mol}\)) + (6 × 16.00 \(\mathrm{g/mol}\)) = 176.13 \(\mathrm{g/mol}\) 2. Calculate the mass of hydrogen in the compound. Mass of hydrogen = 7 × 1.01 \(\mathrm{g/mol}\) = 7.07 \(\mathrm{g/mol}\) 3. Calculate the percentage by mass of hydrogen. Percentage by mass = (7.07 / 176.13) × 100 = 4.01%

(c) Hydrogen in ammonium sulfate, \(\left(\mathrm{NH}_{4}\right)_{2}\mathrm{SO}_{4}\)

1. Calculate the molar mass of ammonium sulfate: \(\left(\mathrm{NH}_{4}\right)_{2}\mathrm{SO}_{4}\) has 2 Nitrogen atoms, 8 Hydrogen atoms, 1 Sulfur atom, and 4 Oxygen atoms. Molar mass = (2 × 14.01 \(\mathrm{g/mol}\)) + (8 × 1.01 \(\mathrm{g/mol}\)) + (1 × 32.07 \(\mathrm{g/mol}\)) + (4 × 16.00 \(\mathrm{g/mol}\)) = 132.14 \(\mathrm{g/mol}\) 2. Calculate the mass of hydrogen in the compound. Mass of hydrogen = 8 × 1.01 \(\mathrm{g/mol}\) = 8.08 \(\mathrm{g/mol}\) 3. Calculate the percentage by mass of hydrogen. Percentage by mass = (8.08 / 132.14) × 100 = 6.11%

(d) Platinum in \(\mathrm{Pt} \mathrm{Cl}_{2}\left(\mathrm{NH}_{3}\right)_{2}\), cisplatin

1. Calculate the molar mass of cisplatin: \(\mathrm{Pt} \mathrm{Cl}_{2}\left(\mathrm{NH}_{3}\right)_{2}\) has 1 Platinum atom, 2 Chlorine atoms, 2 Nitrogen atoms, and 6 Hydrogen atoms. Molar mass = (1 × 195.08 \(\mathrm{g/mol}\)) + (2 × 35.45 \(\mathrm{g/mol}\)) + (2 × 14.01 \(\mathrm{g/mol}\)) + (6 × 1.01 \(\mathrm{g/mol}\)) = 300.06 \(\mathrm{g/mol}\) 2. Calculate the mass of platinum in the compound. Mass of platinum = 1 × 195.08 \(\mathrm{g/mol}\) = 195.08 \(\mathrm{g/mol}\) 3. Calculate the percentage by mass of platinum. Percentage by mass = (195.08 / 300.06) × 100 = 65.01%

(e) Oxygen in estradiol, \(\mathrm{C}_{18} \mathrm{H}_{24} \mathrm{O}_{2}\)

1. Calculate the molar mass of estradiol: \(\mathrm{C}_{18} \mathrm{H}_{24} \mathrm{O}_{2}\) has 18 Carbon atoms, 24 Hydrogen atoms, and 2 Oxygen atoms. Molar mass = (18 × 12.01 \(\mathrm{g/mol}\)) + (24 × 1.01 \(\mathrm{g/mol}\)) + (2 × 16.00 \(\mathrm{g/mol}\)) = 272.39 \(\mathrm{g/mol}\) 2. Calculate the mass of oxygen in the compound. Mass of oxygen = 2 × 16.00 \(\mathrm{g/mol}\) = 32.00 \(\mathrm{g/mol}\) 3. Calculate the percentage by mass of oxygen. Percentage by mass = (32.00 / 272.39) × 100 = 11.75%

(f) Carbon in capsaicin, \(\mathrm{C}_{18}\mathrm{H}_{27}\mathrm{NO}_{3}\)

1. Calculate the molar mass of capsaicin: \(\mathrm{C}_{18}\mathrm{H}_{27}\mathrm{NO}_{3}\) has 18 Carbon atoms, 27 Hydrogen atoms, 1 Nitrogen atom, and 3 Oxygen atoms. Molar mass = (18 × 12.01 \(\mathrm{g/mol}\)) + (27 × 1.01 \(\mathrm{g/mol}\)) + (1 × 14.01 \(\mathrm{g/mol}\)) + (3 × 16.00 \(\mathrm{g/mol}\)) = 305.42 \(\mathrm{g/mol}\) 2. Calculate the mass of carbon in the compound. Mass of carbon = 18 × 12.01 \(\mathrm{g/mol}\) = 216.18 \(\mathrm{g/mol}\) 3. Calculate the percentage by mass of carbon. Percentage by mass = (216.18 / 305.42) × 100 = 70.74%

Key concepts

Molar Mass

Understanding molar mass is fundamental when calculating percentage by mass in chemical compounds. The molar mass of a substance is the weight of one mole (6.022 x 10^23 particles) of that substance. In other words, it's the sum of the atomic masses of all atoms in a molecule, measured in grams per mole (g/mol).

For instance, in the case of acetylene (\textbf{C}\(_2\)\textbf{H}\(_2\)), the molar mass is calculated by adding up the atomic masses of two carbon atoms and two hydrogen atoms. Since each carbon atom's mass is approximately 12.01 g/mol and each hydrogen's is about 1.01 g/mol, the overall molar mass becomes \((2 \times 12.01) + (2 \times 1.01) = 26.04\) g/mol.

It's crucial to find accurate atomic masses for these calculations, typically from the periodic table, ensuring precise molar mass calculation which is the first step in finding percentage by mass. This measurement is the mere foundation of many stoichiometric calculations that greenhorn chemists and students will encounter.

Stoichiometry

Stoichiometry, derived from the Greek words 'stoicheion' (element) and 'metron' (measure), is a branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. This concept helps to predict the amounts of substances consumed and produced during reactions.

To exemplify stoichiometry in terms of percentage by mass, consider acetylene – a simple hydrocarbon. Here, stoichiometry helps in determining the fraction of the compound's mass accounted for by each element. Knowledge of molar masses allows us to relate the mass of individual elements to the overall mass of the compound, resulting in the percentage by mass calculation.

Stoichiometry is not only about balancing chemical equations but understanding the ratios of substances involved. It involves a meticulous approach to chemical calculations and demand for attention to detail, ensuring that measurements adhere to the laws of conservation of mass and proportion.

Empirical Formula

The empirical formula of a chemical compound is a notation that shows the simplest whole-number ratio of elements present in the compound. It's a standardized expression of the compound's composition, not necessarily reflecting the actual number of atoms, but rather the relative number of atoms of each element.

For example, hydrogen peroxide's molecular formula is \textbf{H}\(_2\)\textbf{O}\(_2\), but its empirical formula is \textbf{HO}, indicating that for every atom of oxygen, there's one atom of hydrogen. Similarly, the empirical formula of glucose is \textbf{CH}\(_2\)\textbf{O}, even though its molecular formula is \textbf{C}\(_6\)\textbf{H}\(_{12}\)\textbf{O}\(_6\), speaking to the 1:2:1 ratio of carbon to hydrogen to oxygen.

Understanding the empirical formula is also crucial when delineating the percentage by mass. Once you determine the molar mass and quantify the mass of each element within a compound, you can deduce the simplest whole-number ratio and ultimately the empirical formula. This is often the bridging step between the quantitative analysis of a compound and establishing its structural identity.