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) \(\text{KClO}_{3}\) (b) \(\text{KHSO}_{4}\) (c) \(\text{Na}_{2}\text{CrO}_{4}\)

Step-by-step solution

- Calculate Molar Mass of \(\text{NaClO}_{3}\)

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

- Calculate Percent by Mass of Chlorine in \(\text{NaClO}_{3}\)

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

- Calculate Molar Mass of \(\text{KClO}_{3}\)

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

- Calculate Percent by Mass of Chlorine in \(\text{KClO}_{3}\)

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

- Compare Percent by Mass of Chlorine

Comparing the calculated percentages: \(\text{NaClO}_{3}\) has a higher percent by mass of chlorine (33.3\%) than \(\text{KClO}_{3}\) (28.9\%).

- Calculate Molar Mass of \(\text{KHSO}_{4}\)

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

- Calculate Percent by Mass of Sulfur in \(\text{KHSO}_{4}\)

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

- Calculate Molar Mass of \(\text{K}_{2}\text{SO}_{4}\)

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

- Calculate Percent by Mass of Sulfur in \(\text{K}_{2}\text{SO}_{4}\)

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

- Compare Percent by Mass of Sulfur

Comparing the calculated percentages: \(\text{KHSO}_{4}\) has a higher percent by mass of sulfur (23.5\%) than \(\text{K}_{2}\text{SO}_{4}\) (18.4\%).

- Calculate Molar Mass of \(\text{Na}_{2}\text{CrO}_{4}\)

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

- Calculate Percent by Mass of Chromium in \(\text{Na}_{2}\text{CrO}_{4}\)

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

- Calculate Molar Mass of \(\text{Na}_{2}\text{Cr}_{2}\text{O}_{7}\)

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

- Calculate Percent by Mass of Chromium in \(\text{Na}_{2}\text{Cr}_{2}\text{O}_{7}\)

Percent by mass of chromium in \(\text{Na}_{2}\text{Cr}_{2}\text{O}_{7}\) = \[ \frac{104.00}{261.98} \times 100 \approx 39.7\% \]

- Compare Percent by Mass of Chromium

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

Key concepts

Molar Mass Calculation

To find the molar mass of a compound, you sum the atomic masses of all the atoms in its chemical formula. Each element's atomic mass can be found on the periodic table. For instance, to calculate the molar mass of \(\text{NaClO}_{3}\), you add the atomic mass of sodium (Na), chlorine (Cl), and three oxygen (O) atoms together:
\[ \text{Na} = 22.99, \text{Cl} = 35.45, \text{O}_{3} = 3 \times 16.00 \] Hence, the molar mass of \(\text{NaClO}_{3}\) is:
\[ 22.99 + 35.45 + (3 \times 16.00) = 106.44 \text{ g/mol} \]
After calculating the molar mass, you can proceed to the next steps, such as determining the percent composition by mass of specific elements within the compound.

Percent by Mass

The percent by mass of an element in a compound tells you what fraction of the compound's total mass comes from that particular element. It's expressed as a percentage. To calculate it, you use the formula:
\[ \text{\text{Percent by mass of element X}} = \frac{\text{Mass of X in 1 mole of compound}}{\text{Molar mass of compound}} \times 100 \]
Let's apply this to find the percent by mass of chlorine in \(\text{NaClO}_{3}\):
- Calculate the mass of chlorine in 1 mole of \(\text{NaClO}_{3}\): 35.45 g
- Know the molar mass of \(\text{NaClO}_{3}\): 106.44 g/mol
Thus, the percent by mass of chlorine is:
\[ \frac{\text{35.45}}{\text{106.44}} \times 100 \approx 33.3 \% \]
This means 33.3% of the mass of \(\text{NaClO}_{3}\) comes from chlorine. Repeat similar steps to find the percent by mass for other elements in different compounds.

Elemental Analysis

Elemental analysis involves determining the composition of a chemical compound in terms of the elements present and their amounts. This requires both the molar masses and the percent by mass calculations, as discussed earlier. By comparing percentages of different elements, we can understand how the composition varies across compounds.
For example, in an exercise asking which compound has a lower percent by mass of chlorine, we calculated:
- For \(\text{NaClO}_{3}\), the percent by mass of chlorine was 33.3%
- For \(\text{KClO}_{3}\), it was 28.9%
Using these results, we deduced that \(\text{KClO}_{3}\) has a lower percent by mass of chlorine than \(\text{NaClO}_{3}\). Similar analyses can compare the compositions of sulfur in \(\text{KHSO}_{4}\) and \(\text{K}_{2}\text{SO}_{4}\), or chromium in \(\text{Na}_{2}\text{CrO}_{4}\) and \(\text{Na}_{2}\text{Cr}_{2}\text{O}_{7}\).