Q146CP - Chemistry: Molecular Nature Of Matter And Change

Question

Citric acid (right) is concentrated in citrus fruits and plays a central metabolic role in nearly every animal and plant cell. (a) What are the molar mass and formula of citric acid? (b) How many moles of citric acid are in 1.50 qt of lemon juice (d = 1.09 g/mL) that is 6.82% citric acid by mass?

Step-by-Step Solution

Answer

1. The formula of citric acid is ${\text{C}}_{\text{6}}{\text{H}}_{\text{8}}{\text{O}}_{\text{7}}$and the molar mass is 192.124 g/mol.
2. The moles of citric acid is1.75 mol.

Step 1: Determine the formula of citric acid

The pink dots in the given figure represents the $O$ atom. So, the number of $O$ atoms in citric acid is 7.

The black dot represents the $C$atom, thus the number of $C$atoms in citric acid is 6. And the blue dots represent the $\text{H}$ atom, so the number of $\text{H}$atoms in citric acid is 8.

Therefore, the formula of citric acid is $C_6{H}_8{O}_7.$

Step 2: Calculate the molar mass of the citric acid

The molar mass of = 12.01 g/mol

The molar mass of = 1.008 g/mol

The molar mass of = 16 g/mol

The molar mass of citric acid is calculated as:

$\begin{array}{rcl}molar mass of C_6{H}_8{O}_7& =& \left(6\times molar mass of C\right)+\left(8\times molar mass of H\right)+\left(7\times molar mass of O\right)\\ & & \left(6\times 12.01 g/mol\right)+\left(8\times 1.008 g/mol\right)+\left(7\times 16 g/mol\right)\\ & & 192.124 g/mol\end{array}$

Step 3: Calculate the mass of the citric acid

Since 1 qt is nearly equal to 946.35 mL.

The mass fraction of citric acid(W)  = 6.82 %

The density of citric acid (d) = 1.09 g/mL

The volume of citric acid(V)  = 1.50 qt

Thus, the mass of citric acid(m) is calculated as:

$\begin{array}{rcl}m& =& w d V\\ & =& 0.0682\times 1.09 g/mL\times 4.79\times 946.35 mL\\ & =& 336.98 g\end{array}$

Step 4: Calculate the mass of the citric acid

The molar mass of citric acid = 192.124 g/mol

The mass of citric acid = 336.98 g

The number of moles of citric acid is calculated as:

$\begin{array}{rcl}& & \text{moles} \text{of} \text{citric} \text{acid=}\frac{\text{mass} \text{of} \text{citric} \text{acid}}{\text{Molar} \text{mass} \text{of} \text{citric} \text{acid}}\\ & & \text{ =}\frac{\text{336.98} \text{g}}{\text{192.124} \text{g/mol}}\\ & & \text{ =1.75} \text{mol}\end{array}$