Question

What is the percent composition of Ca(C2H3O2)2?

Select one:
a. 25.3% Ca, 30.4% C, 3.8% H, 40.5% O
b. 40.4% Ca, 24.2% C, 3.1% H, 32.3% O
c. 6.67% Ca, 26.67% C, 40.0% H, 26.67% O
d. 12.5% Ca, 25% C, 37.5% H, 25% O

Answer 1

a

Step-by-step explanation:

relative formula mass of Ca(C2H3O2)2=158

% for Ca=25.3

% for C=30.4

% for H=3.8

% for O=40.5

Answer 2

Choice a. The compound
`\rm Ca(C_2 H_3 O_2)_2` contains approximately
`25.3\%\; \rm Ca`, by mass, approximately
`30.4\%\; \rm C` by mass, approximately
`3.8\%\; \rm H` by mass, and approximately
`40.5\%\; \rm O` by mass.

Step-by-step explanation:

Look up the relative atomic mass of these elements on a modern periodic table:

• 
  `\rm Ca`:
  `40.078`.
• 
  `\rm C`:
  `12.011`.
• 
  `\rm H`:
  `1.008`.
• 
  `\rm O`:
  `15.999`.

The relative atomic mass of an element gives the numerical value of the mass (in grams) of one mole of the atoms of this element. For example, the relative atomic mass of hydrogen is
`1.008`. Therefore, the mass of one mole of hydrogen atoms would be (approximately)
`1.008\; \rm g`.

One mole of
`\rm Ca(C_2 H_3 O_2)_2` formula units include:

• 
  `1\; \rm mol` of
  `\rm Ca` atoms, and
• 
  `2\; \rm mol` of
  `\rm (C_2H_3O_2)^(-)` ions.

Inside each mole of
`\rm (C_2H_3O_2)^(-)` ions, there are:

• 
  `2\; \rm mol` of
  `\rm C` atoms,
• 
  `3\; \rm mol` of
  `\rm H` atoms, and
• 
  `2\; \rm mol` of
  `\rm O` atoms.

Therefore, that
`2\; \rm mol` of
`\rm (C_2H_3O_2)^(-)` ions in one mole of
`\rm Ca(C_2 H_3 O_2)_2` formula units would include:

• 
  `2\; \rm mol * 2 = 4\; \rm mol` of
  `\rm C` atoms,
• 
  `3\; \rm mol * 2 = 6\; \rm mol` of
  `\rm H` atoms, and
• 
  `2\; \rm mol * 2 = 4\; \rm mol` of
  `\rm O` atoms.

In summary, each mole of
`\rm Ca(C_2 H_3 O_2)_2` formula units would contain:

• 
  `1\; \rm mol` of
  `\rm Ca` atoms,
• 
  `4\; \rm mol` of
  `\rm C` atoms,
• 
  `6\; \rm mol` of
  `\rm H` atoms, and
• 
  `4\; \rm mol` of
  `\rm O` atoms.

Calculate the mass of these atoms using their relative atomic mass data.

For example, the mass of one mole of
`\rm H` atoms is (approximately)
`1.008\; \rm g` because the relative atomic mass of hydrogen is
`1.008`. Therefore, the mass of that
`6\; \rm mol` of hydrogen atoms would be approximately:

`6\; \rm mol * 1.008\; \rm g \cdot mol^(-1) = 6.048\; \rm g`.

Similarly, calculate the mass of the other three elements in one mole of
`\rm Ca(C_2 H_3 O_2)_2` formula units:

• 
  `40.078\; \rm g` of
  `\rm Ca`.
• 
  `48.044\; \rm g` of
  `\rm C`.
• 
  `6.048\; \rm g` of
  `\rm H`.
• 
  `63.996\; \rm g` of
  `\rm O`.

The sum of these masses is
`158.166\; \rm g`.

In other words, the mass of one mole of
`\rm Ca(C_2 H_3 O_2)_2` formula units is (approximately)
`158.166\; \rm g`. Approximately
`40.078\; \rm g` of that is
`\rm Ca`. Therefore, the percentage of
`\rm Ca\!` by mass in
`\rm Ca(C_2 H_3 O_2)_2\!` would be approximately:

`\begin{aligned}&100\%* (40.078\; \rm g)/(158.166\; \rm g) \quad \genfrac{}{}{0pt}{}{\leftarrow\text{mass of $\mathrm{Ca}$ in $\mathrm{Ca(C_2 H_3 O_2)_2}$}}{\leftarrow\text{mass of $\mathrm{Ca(C_2 H_3 O_2)_2}$}}\\ & \approx 25.3\% \end{aligned}`.

In other words,
`\rm Ca` accounts for approximately
`25.3\%` of the mass of
`\rm Ca(C_2 H_3 O_2)_2\!`.

In a similar way, calculate the percentage of the other three elements by mass.

• 
  `\rm C`: approximately
  `30.4\%`.
• 
  `\rm H`: approximately
  `3.8\%`.
• 
  `\rm O`: approximately
  `40.5\%`.