Question

If you had 325 grams of a 5.0% by mass solution of citric acid, how many grams would be the acid and how many grams would be water? How many moles of citric acid would it contain?

Answer 1

This solution contains `"16.25 g citric acid"` and `"309 g water"`.
There is `"0.08458 mol"` of citric acid in the solution.

Explanation:

`"mass percent"="grams of solute"/"grams of solution"xx100"`

The mass in grams of a mass % solution, is the sum of the mass in grams of the solute and the mass in grams of the solvent. The solute in this question is citric acid, and the solvent is water.

You need to first calculate the mass in grams of citric acid in `"325 g of solution"`. Then, subtract the mass of the solute from the mass of the solution to get the mass of solvent. Then determine the moles citric acid in the solution.

`color(blue)("Mass of Citric Acid (Solute)"`

Rearrange the mass % equation to isolate grams of solute of citric acid. Insert the known information into the equation and solve.

`"g solute"=("mass percent"xx"mass of solution")/100`

`"g citric acid"=(5.0color(red)cancelcolor(black)("%")xx"325 g solution")/color(red)cancelcolor(black)("100")="16.25 g citric acid"`

`color(blue)("Mass of Water (Solvent)"`

Now subtract the mass in grams of citric acid from the mass in grams of the solution.

`"325 g solution" - "16.25 g citric acid"="309 g water"`

`color(blue)("Moles of Citric Acid"`

Divide the mass of citric acid by its molar mass by multiplying by its inverse.

The molecular formula for citric acid is `"C"_6"H"_8"O"_7`. Its molar mass is `"192.123 g/mol"`. https://www.ncbi.nlm.nih.gov/pccompound?term=%22citric+acid%22

`16.25color(red)cancelcolor(black)("g C"_6"H"_8"O"_7)xx(1"mol C"_6"H"_8"O"_7)/(192.123color(red)cancelcolor(black)("mol C"_6"H"_8"O"_7))="0.08458 mol C"_6"H"_8"O"_7` rounded to two significant figures