Question

Calculate the percent by mass of each element in the following compounds. a. \(\mathrm{HClO}_{3}\) b. UF \(_{4}\) c. \(\mathrm{CaH}_{2}\) d. \(\mathrm{Ag}_{2} \mathrm{~S}\) e. \(\mathrm{NaHSO}_{3}\) f. \(\mathrm{MnO}_{2}\)

Short answer

The percent by mass of each element in the compounds is: a. HClO₃: H = 1.20%, Cl = 41.95%, O = 56.85% b. UF₄: U = 75.87%, F = 24.13% c. CaH₂: Ca = 95.25%, H = 4.75% d. Ag₂S: Ag = 87.02%, S = 12.98% e. NaHSO₃: Na = 22.07%, H = 0.97%, S = 30.78%, O = 46.18% f. MnO₂: Mn = 63.17%, O = 36.83%

Step-by-step solution

Identify the elements in each compound

For each compound, identify the elements present: a. HClO₃: Hydrogen (H), Chlorine (Cl), and Oxygen (O) b. UF₄: Uranium (U) and Fluorine (F) c. CaH₂: Calcium (Ca) and Hydrogen (H) d. Ag₂S: Silver (Ag) and Sulfur (S) e. NaHSO₃: Sodium (Na), Hydrogen (H), Sulfur (S), and Oxygen (O) f. MnO₂: Manganese (Mn) and Oxygen (O)

Determine the molar mass of each compound

Using the periodic table, find the atomic mass of each element in the compound. Multiply by the number of atoms of that element and add the values for all elements to get the molar mass of the compound. a. HClO₃: 1 H + 1 Cl + 3 O = \( (1 \times 1.01) + (1 \times 35.45) + (3 \times 16) = 84.46 \mathrm{g/mol}\) b. UF₄: 1 U + 4 F = \( (1 \times 238.03) + (4 \times 19) = 314.03 \mathrm{g/mol}\) c. CaH₂: 1 Ca + 2 H = \( (1 \times 40.08) + (2 \times 1.01) = 42.10 \mathrm{g/mol}\) d. Ag₂S: 2 Ag + 1 S = \( (2 \times 107.87) + (1 \times 32.07) = 247.81 \mathrm{g/mol}\) e. NaHSO₃: 1 Na + 1 H + 1 S + 3 O = \( (1 \times 22.99) + (1 \times 1.01) + (1 \times 32.07) + (3 \times 16) = 104.07 \mathrm{g/mol}\) f. MnO₂: 1 Mn + 2 O = \( (1 \times 54.94) + (2 \times 16) = 86.94 \mathrm{g/mol}\)

Calculate the percent by mass of each element

For each compound, divide the mass of one element in one mole by the molar mass of the compound and multiply by 100. Repeat this for all elements in the compound. a. HClO₃: H: \(\frac{1.01}{84.46} \times 100 = 1.20 \% \) Cl: \(\frac{35.45}{84.46} \times 100 = 41.95 \% \) O: \(\frac{3 \times 16}{84.46} \times 100 = 56.85 \% \) b. UF₄: U: \(\frac{238.03}{314.03} \times 100 = 75.87 \% \) F: \(\frac{4 \times 19}{314.03} \times 100 = 24.13 \% \) c. CaH₂: Ca: \(\frac{40.08}{42.10} \times 100 = 95.25 \% \) H: \(\frac{2 \times 1.01}{42.10} \times 100 = 4.75 \% \) d. Ag₂S: Ag: \(\frac{2 \times 107.87}{247.81} \times 100 = 87.02 \% \) S: \(\frac{32.07}{247.81} \times 100 = 12.98 \% \) e. NaHSO₃: Na: \(\frac{22.99}{104.07} \times 100 = 22.07 \% \) H: \(\frac{1.01}{104.07} \times 100 = 0.97 \% \) S: \(\frac{32.07}{104.07} \times 100 = 30.78 \% \) O: \(\frac{3 \times 16}{104.07} \times 100 = 46.18 \% \) f. MnO₂: Mn: \(\frac{54.94}{86.94} \times 100 = 63.17 \% \) O: \(\frac{2 \times 16}{86.94} \times 100 = 36.83 \% \) The percent by mass of each element in the compounds is as follows: a. HClO₃: H = 1.20%, Cl = 41.95%, O = 56.85% b. UF₄: U = 75.87%, F = 24.13% c. CaH₂: Ca = 95.25%, H = 4.75% d. Ag₂S: Ag = 87.02%, S = 12.98% e. NaHSO₃: Na = 22.07%, H = 0.97%, S = 30.78%, O = 46.18% f. MnO₂: Mn = 63.17%, O = 36.83%

Key concepts

Molar Mass Calculation

Calculating molar mass is an essential skill in chemistry that helps us determine the weight of one mole of a compound. It is the sum of the masses of all the elements present in a given chemical formula. Here’s how you can do it step by step:

• First, find the atomic mass of each element in the periodic table. This value is usually given in atomic mass units (amu) or grams per mole (g/mol).
• Identify the number of atoms of each element in the compound. You can get this from the subscripts in the chemical formula.
• Multiply the atomic mass of each element by the number of atoms of that element in the formula.
• Finally, add these values together to get the total molar mass of the compound.

For example, if we consider \(\mathrm{HClO}_{3}\), the molar mass is calculated as follows:

• Hydrogen (H): ```1.01 \, g/mol``` x 1 = ```1.01 \, g/mol```
• Chlorine (Cl): ```35.45 \, g/mol``` x 1 = ```35.45 \, g/mol```
• Oxygen (O): ```16 \, g/mol``` x 3 = ```48 \, g/mol```

Adding these up, the total molar mass of \(\mathrm{HClO}_{3}\) is ```84.46 \, g/mol```.

Once you have the molar mass, you can easily move on to other calculations, like finding the percent composition, which is also a critical task in chemistry.

Element Identification

Identifying the elements present in a compound is the foundational step in any chemical analysis. By simply looking at the chemical formula, you can determine all the elements that make up the compound. Here's how to do it:

• The chemical formula uses element symbols from the periodic table. Each symbol corresponds to a different element.
• Subscripts or numbers next to the symbols indicate the quantity of each element in the compound.
• With these symbols and numbers, one can list out all the elements present.

For example, in the compound \(\mathrm{Ag}_{2}\mathrm{~S}\), we have:

• Silver (Ag)
• Sulfur (S)

Here, the subscript '2' next to Ag means there are two atoms of silver in the compound. The absence of a number next to sulfur implies there is only one atom of sulfur present.

Accurate element identification is crucial as it sets the stage for further calculations like molar mass and percent composition of a compound.

Chemical Formulas

Chemical formulas are representations of molecules and compounds that show the types and numbers of atoms involved. They provide us with essential information to understand the composition of the compound.

Different types of chemical formulas serve various purposes, such as:

• Empirical Formula: Shows the simplest whole-number ratio of elements in a compound. Example: \(\mathrm{CH}\) for benzene.
• Molecular Formula: Indicates the actual number of each type of atom present in a molecule. Example: \(\mathrm{C}_6\mathrm{H}_6\) for benzene.

Understanding chemical formulas is key for carrying out precise calculations. They allow us to calculate molar masses and determine the percentage composition of each element in a compound. Let's look at an example formula, \(\mathrm{MnO}_2\), where:

• Manganese (Mn) and Oxygen (O) make up the compound.
• The numbers indicate that the compound consists of one manganese atom and two oxygen atoms.

These informative symbols and numbers enable chemists to decode the composition of compounds efficiently, leading to more effective analysis and experimentation in the laboratory.