Question

Calculate the percent by mass of the element listed first in the formulas for each of the following compounds. a. copper(II) bromide, \(\mathrm{CuBr}_{2}\) b. copper(I) bromide, \(\mathrm{CuBr}\) c. iron(II) chloride, \(\mathrm{FeCl}_{2}\) d. iron(III) chloride, \(\mathrm{FeCl}_{3}\) e. cobalt(II) iodide, \(\mathrm{CoI}_{2}\) f. cobalt(III) iodide, \(\mathrm{CoI}_{3}\) g. tin(II) oxide, \(\mathrm{SnO}\) h. tin(IV) oxide, \(\mathrm{SnO}_{2}\)

Short answer

The percent by mass of the metal element in each compound is: a. copper(II) bromide, \(\mathrm{CuBr}_{2}\) = 28.45% b. copper(I) bromide, \(\mathrm{CuBr}\) = 44.31% c. iron(II) chloride, \(\mathrm{FeCl}_{2}\) = 44.05% d. iron(III) chloride, \(\mathrm{FeCl}_{3}\) = 34.43% e. cobalt(II) iodide, \(\mathrm{CoI}_{2}\) = 18.84% f. cobalt(III) iodide, \(\mathrm{CoI}_{3}\) = 13.40% g. tin(II) oxide, \(\mathrm{SnO}\) = 88.12% h. tin(IV) oxide, \(\mathrm{SnO}_{2}\) = 78.76%

Step-by-step solution

Find molar mass of Copper(II) bromide

Cu: 63.55 g/mol, Br: 79.90 g/mol Molar mass of \(\mathrm{CuBr}_{2}\) = 1 × 63.55 + 2 × 79.90 = 223.35 g/mol

Calculate percent by mass of Cu in \(\mathrm{CuBr}_{2}\)

Percent by mass of Cu = (63.55/223.35) × 100 = 28.45% b. copper(I) bromide, \(\mathrm{CuBr}\)

Find molar mass of Copper(I) bromide

Cu: 63.55 g/mol, Br: 79.90 g/mol Molar mass of \(\mathrm{CuBr}\) = 63.55 + 79.90 = 143.45 g/mol

Calculate percent by mass of Cu in \(\mathrm{CuBr}\)

Percent by mass of Cu = (63.55/143.45) × 100 = 44.31% c. iron(II) chloride, \(\mathrm{FeCl}_{2}\)

Find molar mass of Iron(II) chloride

Fe: 55.85 g/mol, Cl: 35.45 g/mol Molar mass of \(\mathrm{FeCl}_{2}\) = 55.85 + 2 × 35.45 = 126.75 g/mol

Calculate percent by mass of Fe in \(\mathrm{FeCl}_{2}\)

Percent by mass of Fe = (55.85/126.75) × 100 = 44.05% d. iron(III) chloride, \(\mathrm{FeCl}_{3}\)

Find molar mass of Iron(III) chloride

Fe: 55.85 g/mol, Cl: 35.45 g/mol Molar mass of \(\mathrm{FeCl}_{3}\) = 55.85 + 3 × 35.45 = 162.20 g/mol

Calculate percent by mass of Fe in \(\mathrm{FeCl}_{3}\)

Percent by mass of Fe = (55.85/162.20) × 100 = 34.43% e. cobalt(II) iodide, \(\mathrm{CoI}_{2}\)

Find molar mass of Cobalt(II) iodide

Co: 58.93 g/mol, I: 126.90 g/mol Molar mass of \(\mathrm{CoI}_{2}\) = 58.93 + 2 × 126.90 = 312.73 g/mol

Calculate percent by mass of Co in \(\mathrm{CoI}_{2}\)

Percent by mass of Co = (58.93/312.73) × 100 = 18.84% f. cobalt(III) iodide, \(\mathrm{CoI}_{3}\)

Find molar mass of Cobalt(III) iodide

Co: 58.93 g/mol, I: 126.90 g/mol Molar mass of \(\mathrm{CoI}_{3}\) = 58.93 + 3 × 126.90 = 439.63 g/mol

Calculate percent by mass of Co in \(\mathrm{CoI}_{3}\)

Percent by mass of Co = (58.93/439.63) × 100 = 13.40% g. tin(II) oxide, \(\mathrm{SnO}\)

Find molar mass of Tin(II) oxide

Sn: 118.71 g/mol, O: 16.00 g/mol Molar mass of \(\mathrm{SnO}\) = 118.71 + 16.00 = 134.71 g/mol

Calculate percent by mass of Sn in \(\mathrm{SnO}\)

Percent by mass of Sn = (118.71/134.71) × 100 = 88.12% h. tin(IV) oxide, \(\mathrm{SnO}_{2}\)

Find molar mass of Tin(IV) oxide

Sn: 118.71 g/mol, O: 16.00 g/mol Molar mass of \(\mathrm{SnO}_{2}\) = 118.71 + 2 × 16.00 = 150.71 g/mol

Calculate percent by mass of Sn in \(\mathrm{SnO}_{2}\)

Percent by mass of Sn = (118.71/150.71) × 100 = 78.76%

Key concepts

Molar Mass Calculation

When you are working with chemical compounds, understanding how to find the molar mass is crucial. The molar mass is the weight of one mole (6.022 x 10²³ particles) of a substance. You can think of it as the molecule's weight in grams.
This calculation is necessary for activities like finding the percent by mass in a compound. To calculate the molar mass, follow these steps:

• Identify all elements in the compound.
• Find the atomic mass of each element on the periodic table. This value is usually below an element's symbol and represents the weight of one mole of atoms in grams.
• Multiply the atomic mass by the number of atoms of that element in the compound.
• Add these values for all components to find the total molar mass of the compound.

Understanding molar mass calculation forms the backbone of many chemical computations. By mastering this simple yet essential skill, you can tackle more complex chemical problems with confidence.

Chemical Compounds

Chemical compounds are made up of atoms that are bonded together. These atoms can belong to the same element or different elements. The way that these atoms bond can affect the compound's properties, like its state of matter, color, and reactivity.
Chemical compounds are often represented by chemical formulas, which tell us which atoms are present and in what quantity. For example, water has the formula \(\text{H}_2\text{O}\). This means each water molecule contains two hydrogen atoms and one oxygen atom.

• Each element is represented by its atomic symbol, such as H for hydrogen or Cl for chlorine.
• The subscript numbers in a formula indicate how many of each element's atoms are present. If no number is shown, it means one atom is present.
• Understanding this notation helps in calculating other properties of the compound, like its molar mass or percent composition.

Through exploring the world of chemical compounds, you gain insights into how substances interact, form, and change in chemistry.

Elemental Analysis

Elemental analysis involves determining the quantity of each element present in a compound. This method is vital for elucidating a compound's chemical formula and studying its composition. One common form of elemental analysis is calculating the percent composition of a compound.
The percent composition gives us the relative percent of each element in a compound based on its molar mass.

• To find the percent mass of an element, divide the total mass of the element in the compound by the compound's total molar mass, then multiply by 100.
• This value helps chemists understand how elements are distributed within the compound.
• Percent composition can also guide synthesis and analysis of compounds in labs.

By mastering elemental analysis, you can unlock a deeper understanding of chemical compositions and refine your approach to chemical experimentation and studies.