Question

Examine the following formulas. Which compound has the (a) lower percent by mass of chlorine: \(\mathrm{NaClO}_{3}\) or \(\mathrm{KClO}_{3}\) ? (b) higher percent by mass of sulfur: \(\mathrm{KHSO}_{4}\) or \(\mathrm{K}_{2} \mathrm{SO}_{4}\) ? (c) lower percent by mass of chromium: \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) or \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) ?

Short answer

(a) KClO₃ (b) KHSO₄ (c) Na₂CrO₄

Step-by-step solution

Calculate Molar Mass - NaClO₃

Find the molar mass of \(\text{NaClO}_{3}\). Molar masses are: Na = 22.99 g/mol, Cl = 35.45 g/mol, and O = 16.00 g/mol. \[ \text{Molar mass of NaClO}_{3} = 22.99 + 35.45 + 3 \times 16.00 = 106.44 \text{ g/mol} \]

Calculate Percent Mass of Chlorine in NaClO₃

Calculate the percent by mass of chlorine in \(\text{NaClO}_{3}\): \[ \text{Percent by mass of Cl} = \frac{35.45}{106.44} \times 100 = 33.3\text{%} \]

Calculate Molar Mass - KClO₃

Find the molar mass of \(\text{KClO}_{3}\). Molar masses are: K = 39.10 g/mol, Cl = 35.45 g/mol, and O = 16.00 g/mol. \[ \text{Molar mass of KClO}_{3} = 39.10 + 35.45 + 3 \times 16.00 = 122.55 \text{ g/mol} \]

Calculate Percent Mass of Chlorine in KClO₃

Calculate the percent by mass of chlorine in \(\text{KClO}_{3}\): \[ \text{Percent by mass of Cl} = \frac{35.45}{122.55} \times 100 = 28.9\text{%} \]

Compare Percent Mass of Chlorine

Compare the percent masses calculated: \(\text{NaClO}_{3}: 33.3\text{%}\) and \(\text{KClO}_{3}: 28.9\text{%}\). \(\text{KClO}_{3}\) has the lower percent by mass of chlorine.

Calculate Molar Mass - KHSO₄

Find the molar mass of \(\text{KHSO}_{4}\). Molar masses are: K = 39.10 g/mol, H = 1.01 g/mol, S = 32.07 g/mol, and O = 16.00 g/mol. \[ \text{Molar mass of KHSO}_{4} = 39.10 + 1.01 + 32.07 + 4 \times 16.00 = 136.18 \text{ g/mol} \]

Calculate Percent Mass of Sulfur in KHSO₄

Calculate the percent by mass of sulfur in \(\text{KHSO}_{4}\): \[ \text{Percent by mass of S} = \frac{32.07}{136.18} \times 100 = 23.5\text{%} \]

Calculate Molar Mass - K₂SO₄

Find the molar mass of \(\text{K}_{2}\text{SO}_{4}\). Molar masses are: K = 39.10 g/mol, S = 32.07 g/mol, and O = 16.00 g/mol. \[ \text{Molar mass of K}_{2}\text{SO}_{4} = 2 \times 39.10 + 32.07 + 4 \times 16.00 = 174.27 \text{ g/mol} \]

Calculate Percent Mass of Sulfur in K₂SO₄

Calculate the percent by mass of sulfur in \(\text{K}_{2}\text{SO}_{4}\): \[ \text{Percent by mass of S} = \frac{32.07}{174.27} \times 100 = 18.4\text{%} \]

Compare Percent Mass of Sulfur

Compare the percent masses calculated: \(\text{KHSO}_{4}: 23.5\text{%}\) and \(\text{K}_{2}\text{SO}_{4}: 18.4\text{%}\). \(\text{KHSO}_{4}\) has the higher percent by mass of sulfur.

Calculate Molar Mass - Na₂CrO₄

Find the molar mass of \(\text{Na}_{2}\text{CrO}_{4}\). Molar masses are: Na = 22.99 g/mol, Cr = 52.00 g/mol, and O = 16.00 g/mol. \[ \text{Molar mass of Na}_{2}\text{CrO}_{4} = 2 \times 22.99 + 52.00 + 4 \times 16.00 = 161.98 \text{ g/mol} \]

Calculate Percent Mass of Chromium in Na₂CrO₄

Calculate the percent by mass of chromium in \(\text{Na}_{2}\text{CrO}_{4}\): \[ \text{Percent by mass of Cr} = \frac{52.00}{161.98} \times 100 = 32.1\text{%} \]

Calculate Molar Mass - Na₂Cr₂O₇

Find the molar mass of \(\text{Na}_{2}\text{Cr}_{2}\text{O}_{7}\). Molar masses are: Na = 22.99 g/mol, Cr = 52.00 g/mol, and O = 16.00 g/mol. \[ \text{Molar mass of Na}_{2}\text{Cr}_{2}\text{O}_{7} = 2 \times 22.99 + 2 \times 52.00 + 7 \times 16.00 = 261.98 \text{ g/mol} \]

Calculate Percent Mass of Chromium in Na₂Cr₂O₇

Calculate the percent by mass of chromium in \(\text{Na}_{2}\text{Cr}_{2}\text{O}_{7}\): \[ \text{Percent by mass of Cr} = \frac{2 \times 52.00}{261.98} \times 100 = 39.7\text{%} \]

Compare Percent Mass of Chromium

Compare the percent masses calculated: \(\text{Na}_{2}\text{CrO}_{4}: 32.1\text{%}\) and \(\text{Na}_{2}\text{Cr}_{2}\text{O}_{7}: 39.7\text{%}\). \(\text{Na}_{2}\text{CrO}_{4}\) has the lower percent by mass of chromium.

Key concepts

molar mass

Molar mass is a crucial concept in chemistry. It represents the mass of one mole of a substance, measured in grams per mole (g/mol). To find the molar mass of a compound, you add up the molar masses of all the elements in its chemical formula.
For example, to calculate the molar mass of \(\text{NaClO}_3\), you have:

• Sodium (Na) = 22.99 g/mol
• Chlorine (Cl) = 35.45 g/mol
• Oxygen (O) = 16.00 g/mol (and there are 3 oxygens in the compound)

The molar mass of \(\text{NaClO}_3\) is calculated as:
\text{22.99 + 35.45 + 3 \(\times 16.00\)} = 106.44 g/mol
Understanding molar mass helps in calculating other key properties, such as percent composition and stoichiometry.

chemical formulas

Chemical formulas give the exact number of each type of atom in a molecule. For example, \(\text{H}_2\text{O}\) tells us there are 2 hydrogen atoms and 1 oxygen atom in a water molecule.
They are essential for determining the molar mass and percent composition of a compound. Each element in a formula has an associated atomic mass, which you sum to find the molar mass. In our example, for \(\text{KClO}_3\), the formula tells us there is:

• 1 Potassium (K)
• 1 Chlorine (Cl)
• 3 Oxygens (O)

With these in mind, calculating the molar mass or any other chemical property becomes easier.
Remember, the subscripts in the chemical formulas are vital as they indicate the ratio in which elements combine to form a compound.

percent by mass

Percent by mass, also called percent composition, is the percentage of the total mass of a compound that comes from each element in that compound. It is calculated using the formula:
\[ \text{Percent by mass of Element} = \frac{\text{Mass of Element in 1 mole of compound}}{\text{Molar Mass of compound}} \ \times 100 \] For example, to find the percent by mass of chlorine in \(\text{KClO}_3\):

• Calculate the molar mass: \text{122.55 g/mol}
• Mass of Cl in the compound: \text{35.45 g/mol}

Plug these into the formula:
\[ \text{Percent by mass of Cl} = \frac{\text{35.45}}{\text{122.55}} \ \times 100 = 28.9\text{%} \] This tells you that chlorine makes up 28.9% of the mass of \(\text{KClO}_3\). Understanding percent by mass is vital for characterizing the composition of compounds.