Question

Determine the chemical formula of each compound and use it to calculate the mass percent composition of each constituent element. $$ \begin{array}{l}{\text { a. potassium chromate }} \\ {\text { b. lead(II) phosphate }} \\ {\text { c. cobalt(II) bromide }}\end{array} $$

Short answer

a. K2CrO4: Mass percent of K = 40.18%, Cr = 26.78%, O = 33.04%. b. Pb3(PO4)2: Mass percent of Pb = 76.62%, P = 7.60%, O = 15.78%. c. CoBr2: Mass percent of Co = 26.93%, Br = 73.07%.

Step-by-step solution

Determine the Chemical Formula of Potassium Chromate

To determine the chemical formula of potassium chromate, identify the ions and their charges: potassium (K) has a charge of +1, and the chromate ion (CrO4) has a charge of -2. Since the charges must balance in a neutral compound, two potassium ions are needed for every chromate ion. The formula for potassium chromate is K2CrO4.

Calculate the Mass Percent of Each Element in Potassium Chromate

First, obtain the atomic masses of the elements from the periodic table: K = 39.10 amu, Cr = 51.996 amu, O = 16.00 amu. Then calculate the molar mass: 2\(\times\)39.10 (K) + 51.996 (Cr) + 4\(\times\)16.00 (O) = 194.196 amu. Next, divide the total mass of each element by the molar mass and multiply by 100%: Mass percent of K = \(\frac{2\times 39.10}{194.196}\times 100\%\), Mass percent of Cr = \(\frac{51.996}{194.196}\times 100\%\), Mass percent of O = \(\frac{4\times 16.00}{194.196}\times 100\%\).

Determine the Chemical Formula of Lead(II) Phosphate

Lead(II) phosphate consists of lead (Pb) with a +2 charge and the phosphate ion (PO4) with a -3 charge. Balancing the charges, we get three lead ions for every two phosphate ions. Thus, the chemical formula for lead(II) phosphate is Pb3(PO4)2.

Calculate the Mass Percent of Each Element in Lead(II) Phosphate

Using atomic masses: Pb = 207.2 amu, P = 30.974 amu, O = 16.00 amu, find the molar mass: 3\(\times\)207.2 (Pb) + 2\(\times\)30.974 (P) +8\(\times\)16.00 (O) = 811.516 amu. Calculate the mass percent of each element: Mass percent of Pb = \(\frac{3\times 207.2}{811.516}\times 100\%\), Mass percent of P = \(\frac{2\times 30.974}{811.516}\times 100\%\), Mass percent of O = \(\frac{8\times 16.00}{811.516}\times 100\%\).

Determine the Chemical Formula of Cobalt(II) Bromide

Cobalt(II) has a +2 charge and bromide (Br) has a -1 charge. Two bromide ions will balance out the charge of one cobalt ion, which gives us the formula CoBr2 for cobalt(II) bromide.

Calculate the Mass Percent of Each Element in Cobalt(II) Bromide

Take the atomic masses: Co = 58.933 amu, Br = 79.904 amu. Calculate the molar mass: 58.933 (Co) + 2\(\times\)79.904 (Br) = 218.741 amu. Calculate mass percent: Mass percent of Co = \(\frac{58.933}{218.741}\times 100\%\), Mass percent of Br = \(\frac{2\times 79.904}{218.741}\times 100\%\).

Key concepts

Chemical Formula Calculation

Determining the chemical formula of a compound is a fundamental step in chemistry that requires an understanding of ions and their charges. Elements form ions to become stable, typically by losing or gaining electrons to obtain a full outer shell. Ions have charges, and when they combine to form compounds, their total charges must balance to create a neutrally charged compound.

For example, in potassium chromate (K2CrO4), potassium ions (K+) have a charge of +1, and chromate ions (CrO4 2-) have a charge of -2. To achieve neutrality, the compound requires two potassium ions for every chromate ion, thus we get the chemical formula K2CrO4. This balancing act can be simplified as the crossing-over method, where the magnitude of the charge on one ion becomes the subscript of the other and vice versa.

Identifying the right stoichiometry, which is the ratio of ions within a compound, is critical for accurate chemical formula calculation. Doing so allows us to measure and predict the properties of the substance, including the mass percent composition of each constituent element.

Molar Mass

Once the chemical formula of a compound is established, we can calculate its molar mass, which is the mass of one mole of that substance. The molar mass is typically expressed in grams per mole (g/mol) and is calculated by adding together the atomic masses of all atoms present in a molecule, as determined by the chemical formula.

For instance, the molar mass of potassium chromate (K2CrO4) is obtained by summing twice the atomic mass of potassium, once the atomic mass of chromium, and four times the atomic mass of oxygen. Atomic mass is often measured in atomic mass units (amu), with one mole of a substance being equivalent to its atomic mass in grams. The precision in determining molar mass is essential, as it impacts the calculation of the mass percent composition and other quantitative aspects of stoichiometry and chemistry.

Molar mass serves not only as a bridge between the microscopic world of atoms and molecules and the macroscopic world of grams and liters but also as a crucial factor in chemical reactions for calculating yields and reactant usage.

Atomic Mass

Atomic mass, sometimes referred to as atomic weight, is the mass of a single atom of an element. It is measured in atomic mass units (amu), where one amu is defined as one-twelfth the mass of a carbon-12 atom. The atomic mass of an element is a weighted average of the masses of all the naturally occurring isotopes of that element, and it reflects both the mass and the abundance of each isotope.

The atomic mass of an element can be found on the periodic table and is critical for various calculations in chemistry. When it comes to determining the mass percent composition of a compound, the atomic mass of each element is used to calculate the molar mass, which then serves as the basis for the percentage calculation. For example, the atomic mass of bromine (Br) is 79.904 amu, an average that accounts for the proportions of its stable isotopes, bromine-79 and bromine-81.

Understanding atomic mass is essential since it provides a link between the properties of individual atoms and larger-scale quantities of a substance, enabling scientists and students to perform accurate, quantitative analyses in chemical studies and various applications.