Question

Calculate the percentage of (a) mercury in \(\mathrm{HgCO}_{3}\) (b) oxygen in \(\mathrm{Ca}\left(\mathrm{ClO}_{3}\right)_{2}\) (c) nitrogen in \(\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{~N}_{2}\) (nicotine) (d) \(\mathrm{Mg}\) in \(\mathrm{C}_{55} \mathrm{H}_{72} \mathrm{MgN}_{4} \mathrm{O}_{5}\) (chlorophyll)

Short answer

(a) 76.98% Hg (b) 46.37% O (c) 17.27% N (d) 2.72% Mg

Step-by-step solution

Determine the Molar Mass of Each Compound

Calculate the molar mass for each compound by summing the atomic masses of all atoms in each molecule: (a) \(\text{HgCO}_{3}\): atomic masses: Hg = 200.59, C = 12.01, O = 16.00 (b) \(\text{Ca(ClO}_{3}\text{)}_{2}\): atomic masses: Ca = 40.08, Cl = 35.45, O = 16.00 (c) \(\text{C}_{10}\text{H}_{14}\text{N}_{2}\): atomic masses: C = 12.01, H = 1.01, N = 14.01 (d) \(\text{C}_{55}\text{H}_{72}\text{MgN}_{4}\text{O}_{5}\): atomic masses: C = 12.01, H = 1.01, Mg = 24.31, N = 14.01, O = 16.00

Calculate the Total Molar Mass

(a) \[ \text{Total Molar Mass of HgCO}_{3} = 200.59 + 12.01 + (3 \times 16.00) = 260.60 \] (b) \[ \text{Total Molar Mass of Ca(ClO}_{3}\text{)}_{2} = 40.08 + (2 \times 35.45) + (6 \times 16.00) = 206.98 \] (c) \[ \text{Total Molar Mass of C}_{10}\text{H}_{14}\text{N}_{2} = (10 \times 12.01) + (14 \times 1.01) + (2 \times 14.01) = 162.23 \] (d) \[ \text{Total Molar Mass of C}_{55}\text{H}_{72}\text{MgN}_{4}\text{O}_{5} = (55 \times 12.01) + (72 \times 1.01) + 24.31 + (4 \times 14.01) + (5 \times 16.00) = 893.49 \]

Calculate the Mass Contribution of the Specific Element

(a) For Hg in \(\text{HgCO}_{3}\), mass of Hg is 200.59. (b) For O in \(\text{Ca(ClO}_{3}\text{)}_{2}\), mass of O is (6 \times 16.00) = 96.00. (c) For N in \(\text{C}_{10}\text{H}_{14}\text{N}_{2}\), mass of N is (2 \times 14.01) = 28.02. (d) For Mg in \(\text{C}_{55}\text{H}_{72}\text{MgN}_{4}\text{O}_{5}\), mass of Mg is 24.31.

Calculate the Percentage by Mass

(a) Percentage of Hg in \(\text{HgCO}_{3}\): \[ \frac{200.59}{260.60} \times 100 = 76.98\% \] (b) Percentage of Oxygen in \(\text{Ca(ClO}_{3}\text{)}_{2}\): \[ \frac{96.00}{206.98} \times 100 = 46.37\% \] (c) Percentage of Nitrogen in \(\text{C}_{10}\text{H}_{14}\text{N}_{2}\): \[ \frac{28.02}{162.23} \times 100 = 17.27\% \] (d) Percentage of Mg in \(\text{C}_{55}\text{H}_{72}\text{MgN}_{4}\text{O}_{5}\): \[ \frac{24.31}{893.49} \times 100 = 2.72\% \]

Key concepts

molar mass calculation

To understand the percentage composition of a compound, we need to start by calculating its molar mass. The molar mass of a compound is the sum of the atomic masses of all the atoms in its molecular formula. Here's a simple approach:
1. Identify each element in the compound and note its atomic mass.
2. Multiply the atomic mass of each element by the number of times it appears in the molecule.
3. Sum these values to get the total molar mass.
For example, in \(\mathrm{HgCO\_3}\), we have: Hg = 200.59, C = 12.01, O = 16.00, and we need to account for three O atoms. Therefore, the total molar mass is: \[ 200.59 + 12.01 + (3 \times 16.00) = 260.60 \] Practicing these steps will help ensure accuracy in further calculations.

mass contribution

Once the molar mass is calculated, the next step is to determine the mass contribution of the specific element within the compound. This step is crucial for calculating the percentage composition. Follow these steps:
1. Identify the atomic mass of the element of interest.
2. Multiply it by the number of atoms of that element in the molecular formula.
For \( \mathrm{Ca(ClO\_3)\_2} \), we're interested in the mass of oxygen: We have 6 O atoms, each contributing 16.00 to the molar mass. Therefore, the total mass contribution of oxygen is: \[ 6 \times 16.00 = 96.00 \] Careful attention to detail in calculating these contributions is key to ensuring accurate results.

elemental composition

Finally, using the molar mass and the mass contribution, we calculate the percentage by mass of an element in the compound. The formula for percentage by mass is:
\[ \frac{\text{Mass of the element}}{\text{Total molar mass}} \times 100 \] Let's put this into practice:
For manganese (Mg) in \( \mathrm{C\_55 H\_72 MgN\_4O\_5} \), we have: \
- Mass of Mg: 24.31
- Total molar mass: 893.49
The percentage by mass of Mg is: \[ \frac{24.31}{893.49} \times 100 = 2.72 \% \] When you break down the problem into these steps, calculating percentage compositions becomes straightforward and manageable.