Question

Calculate the percent by mass of the element listed first in the formulas for each of the following compounds. a. iron(III) chloride b. oxygen difluoride, \(\mathrm{OF}_{2}\) c. benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\) d. ammonium perchlorate, \(\mathrm{NH}_{4} \mathrm{ClO}_{4}\) e. silver oxide f. cobalt(II) chloride g. dinitrogen tetroxide h. manganese(II) chloride

Short answer

The percent by mass of the first element in each compound is as follows: a. Iron(III) chloride: 34.4% Fe b. Oxygen difluoride (\(OF_2\)): 29.6% O c. Benzene (\(C_6H_6\)): 92.2% C d. Ammonium perchlorate (\(NH_4ClO_4\)): 11.9% N e. Silver oxide: 93.1% Ag f. Cobalt(II) chloride: 45.4% Co g. Dinitrogen tetroxide: 30.4% N h. Manganese(II) chloride: 43.7% Mn

Step-by-step solution

Write the formula for iron(III) chloride

The formula for iron(III) chloride is FeCl3.

Calculate the total mass of iron and chloride in the compound

We have 1 iron (Fe) atom with a molar mass of 55.85 g/mol and 3 chloride (Cl) atoms with a molar mass of each 35.45 g/mol. Total mass of Fe = 1 * 55.85 g/mol Total mass of Cl = 3 * 35.45 g/mol = 106.35 g/mol Total mass of FeCl3 = 55.85 g/mol + 106.35 g/mol = 162.20 g/mol

Calculate the percent by mass of iron(III) chloride

Percent by mass of iron in FeCl3 = (total mass of Fe / total mass of FeCl3) * 100% Percent by mass = (55.85 g/mol / 162.20 g/mol) * 100% = 34.4% b. Oxygen difluoride, OF2

Write the formula for oxygen difluoride

The formula for oxygen difluoride is OF2.

Calculate the total mass of oxygen and fluorine in the compound

We have 1 oxygen (O) atom with a molar mass of 16.00 g/mol and 2 fluorine (F) atoms with a molar mass of each 19.00 g/mol. Total mass of O = 1 * 16.00 g/mol Total mass of F = 2 * 19.00 g/mol = 38.00 g/mol Total mass of OF2 = 16.00 g/mol + 38.00 g/mol = 54.00 g/mol

Calculate the percent by mass of oxygen difluoride

Percent by mass of oxygen in OF2 = (total mass of O / total mass of OF2) * 100% Percent by mass = (16.00 g/mol / 54.00 g/mol) * 100% = 29.6% c. Benzene, C6H6

Write the formula for benzene

The formula for benzene is C6H6.

Calculate the total mass of carbon and hydrogen in the compound

We have 6 carbon (C) atoms with a molar mass of each 12.01 g/mol and 6 hydrogen (H) atoms with a molar mass of each 1.008 g/mol. Total mass of C = 6 * 12.01 g/mol = 72.06 g/mol Total mass of H = 6 * 1.008 g/mol = 6.048 g/mol Total mass of C6H6 = 72.06 g/mol + 6.048 g/mol = 78.108 g/mol

Calculate the percent by mass of benzene

Percent by mass of carbon in C6H6 = (total mass of C / total mass of C6H6) * 100% Percent by mass = (72.06 g/mol / 78.108 g/mol) * 100% = 92.2% d. Ammonium perchlorate, NH4ClO4

Write the formula for ammonium perchlorate

The formula for ammonium perchlorate is NH4ClO4.

Calculate the total mass of nitrogen, hydrogen, chlorine, and oxygen in the compound

We have 1 nitrogen (N) atom with a molar mass of 14.01 g/mol, 4 hydrogen (H) atoms with a molar mass of each 1.008 g/mol, 1 chlorine (Cl) atom with a molar mass of 35.45 g/mol, and 4 oxygen (O) atoms with a molar mass of each 16.00 g/mol. Total mass of N = 1 * 14.01 g/mol Total mass of H = 4 * 1.008 g/mol = 4.032 g/mol Total mass of Cl = 1 * 35.45 g/mol Total mass of O = 4 * 16.00 g/mol = 64.00 g/mol Total mass of NH4ClO4 = 14.01 g/mol + 4.032 g/mol + 35.45 g/mol + 64.00 g/mol = 117.492 g/mol

Calculate the percent by mass of ammonium perchlorate

Percent by mass of nitrogen in NH4ClO4 = (total mass of N / total mass of NH4ClO4) * 100% Percent by mass = (14.01 g/mol / 117.492 g/mol) * 100% = 11.9% e. Silver oxide

Write the formula for silver oxide

The formula for silver oxide is Ag2O.

Calculate the total mass of silver and oxygen in the compound

We have 2 silver (Ag) atoms with a molar mass of each 107.87 g/mol and 1 oxygen (O) atom with a molar mass of 16.00 g/mol. Total mass of Ag = 2 * 107.87 g/mol = 215.74 g/mol Total mass of O = 1 * 16.00 g/mol Total mass of Ag2O = 215.74 g/mol + 16.00 g/mol = 231.74 g/mol

Calculate the percent by mass of silver oxide

Percent by mass of silver in Ag2O = (total mass of Ag / total mass of Ag2O) * 100% Percent by mass = (215.74 g/mol / 231.74 g/mol) * 100% = 93.1% f. Cobalt(II) chloride

Write the formula for cobalt(II) chloride

The formula for cobalt(II) chloride is CoCl2.

Calculate the total mass of cobalt and chlorine in the compound

We have 1 cobalt (Co) atom with a molar mass of 58.93 g/mol and 2 chlorine (Cl) atoms with a molar mass of each 35.45 g/mol. Total mass of Co = 1 * 58.93 g/mol Total mass of Cl = 2 * 35.45 g/mol = 70.90 g/mol Total mass of CoCl2 = 58.93 g/mol + 70.90 g/mol = 129.83 g/mol

Calculate the percent by mass of cobalt(II) chloride

Percent by mass of cobalt in CoCl2 = (total mass of Co / total mass of CoCl2) * 100% Percent by mass = (58.93 g/mol / 129.83 g/mol) * 100% = 45.4% g. Dinitrogen tetroxide

Write the formula for dinitrogen tetroxide

The formula for dinitrogen tetroxide is N2O4.

Calculate the total mass of nitrogen and oxygen in the compound

We have 2 nitrogen (N) atoms with a molar mass of each 14.01 g/mol and 4 oxygen (O) atoms with a molar mass of each 16.00 g/mol. Total mass of N = 2 * 14.01 g/mol = 28.02 g/mol Total mass of O = 4 * 16.00 g/mol = 64.00 g/mol Total mass of N2O4 = 28.02 g/mol + 64.00 g/mol = 92.02 g/mol

Calculate the percent by mass of dinitrogen tetroxide

Percent by mass of nitrogen in N2O4 = (total mass of N / total mass of N2O4) * 100% Percent by mass = (28.02 g/mol / 92.02 g/mol) * 100% = 30.4% h. Manganese(II) chloride

Write the formula for manganese(II) chloride

The formula for manganese(II) chloride is MnCl2.

Calculate the total mass of manganese and chlorine in the compound

We have 1 manganese (Mn) atom with a molar mass of 54.94 g/mol and 2 chlorine (Cl) atoms with a molar mass of each 35.45 g/mol. Total mass of Mn = 1 * 54.94 g/mol Total mass of Cl = 2 * 35.45 g/mol = 70.90 g/mol Total mass of MnCl2 = 54.94 g/mol + 70.90 g/mol = 125.84 g/mol

Calculate the percent by mass of manganese(II) chloride

Percent by mass of manganese in MnCl2 = (total mass of Mn / total mass of MnCl2) * 100% Percent by mass = (54.94 g/mol / 125.84 g/mol) * 100% = 43.7%

Key concepts

Chemical Compound Composition

Understanding the composition of a chemical compound is fundamental for various fields of chemistry, including pharmaceutical development, materials science, and environmental studies. The composition of a compound is represented by its chemical formula, which indicates the types and numbers of each atom present. For instance, the formula for iron(III) chloride, FeCl₃, implies that it contains one iron (Fe) atom and three chloride (Cl) atoms bonded together.

When analyzing percent by mass calculations, we focus on the individual contributions of these atoms to the overall mass of the compound. Calculating these percentages allows scientists to deduce the purity of a sample, adjust reaction stoichiometry, and prepare solutions with accurate concentrations. Furthermore, understanding the composition on a mass basis aids in regulatory compliance for substances across various industries.

Molar Mass Determination

Molar mass is a pivotal concept in chemistry, representing the mass of one mole (6.022 x 10²³ particles) of a substance. It has units of grams per mole (g/mol) and is crucial when converting between mass and the number of moles in a substance. The molar mass of a compound is the sum of the atomic masses of all the atoms in its formula. For instance, in benzene (C₆H₆), the molar mass is calculated by multiplying the atomic mass of carbon (12.01 g/mol) by 6 and the atomic mass of hydrogen (1.008 g/mol) by 6, then summing these products.

Determining the molar mass allows chemists to measure specific quantities of substances for reactions, which is crucial for the accuracy and reproducibility of experimental results. The knowledge of molar mass is also pertinent when interpreting mass spectra in analytical chemistry, as well as for calculations involving gas laws.

Stoichiometry

Stoichiometry is the quantitative aspect of chemical reactions, describing the relationship between reactants and products. It's comparable to following a recipe - one must use the right amount of each ingredient to achieve the desired outcome. The coefficients in a balanced chemical equation provide the ratios of moles of reactants that react and moles of products formed. For example, in the reaction between silver oxide and hydrogen to form silver and water, the stoichiometry dictates that one mole of silver oxide reacts with one mole of hydrogen to yield two moles of silver and one mole of water.

By harnessing stoichiometry, chemists can predict the amounts of products generated from given reactants, enabling effective resource management and waste minimization. Moreover, it underpins the calculations of chemical yields, and it's essential for scaling reactions from laboratory to industrial scale. The concept is not confined to the laboratory; it's also applied in biotechnology, environmental engineering, and many other disciplines where chemical transformations are pivotal.