# Question a9bde

Jul 16, 2017

3.23%

#### Explanation:

NOTE: I'll assume the compound is sodium bicarbonate, ${\text{NaHCO}}_{3}$, not ${\text{NaHCD}}_{3}$ (even though for this problem it truly doesn't matter).

We're asked to find the percent by mass of ${\text{NaHCO}}_{3}$ given that the masses of solute and water present are $20$ $\text{g}$ and $600$ $\text{g}$ respectively.

To do this, we can use the equation

%"mass solute" = "mass of solute"/"mass of solution" xx 100%

We known the mass of solute is $20$ $\text{g}$.

The mass of the solution is

overbrace(20color(white)(l)"g")^("mass of NaHCO"_3) + overbrace(600color(white)(l)"g")^"mass of water" = color(red)(620 color(red)("g soln"

Therefore, we have

%"mass NaHCO"_3 = (20color(white)(l)"g NaHCO"_3)/(color(red)(620)color(white)(l)color(red)("g solution")) xx 100% = color(blue)(ul(3.23%#