1. Chemical Mixture Problems
Solving
Chemical Mixture Problems
Algebra 1H Glencoe McGraw-Hill J. Evans/C. Logan

2. Mixture problems were introduced earlier this year
Mixture problems were introduced earlier this year. In those problems we saw different solid ingredients like prunes and apricots, each at their own price, combined together to form a mixture at a new price. + =
1st ingredient
2nd ingredient
mixture +
cost · amount
cost · amount =
cost · amount
Chemical mixture problems are another type of mixture problem. Instead of a cost for each ingredient, we’ll consider the strength of the solution, measured in percents.

3. + = First equation: 40 + x = y 40 x y
Ex. 1: Mrs. Armstrong has 40 mL of a solution that is 50% acid. How much water should she add to make a solution that is 10% acid? 40 x y + =
50% ACID SOLUTION
PURE WATER
10% ACID SOLUTION
The problem asks for the amount of water Mrs. Armstrong should add.
Let x = amount of water added
We also need to know how much of the 10% solution she’ll end up with.
Let y = amount of new solution
First equation: x = y

4. The first equation only addressed the amount of the liquids.
40 + x = y
The equation says that Mrs. Armstrong started with 40 mL of a strong acid solution, then added x mL of water and ended up with y mL of a weaker acid solution.
The second equation in the system needs to address the strength of each solution (percentage of acid).

5. Why is the percentage on the water 0%?x y 40 + =
10% ACID SOLUTION
50% ACID SOLUTION
WATER + =
% · amount
% · amount
% · amount
.50 · · x = · y
= .1y
20 = .1y
Why is the percentage on the water 0%?

6. Mrs. Armstrong needs to add 160 mL of water.
Solve the system using any method.
.50(40) + 0x = .10y
= .1y
20 = .1y
200 = y
Mrs. Armstrong needs to add 160 mL of water.

7. - = First equation: 50 - x = y
Ex. 2: How many liters of water must Mr. Wade EVAPORATE from 50 L of a 10% salt solution to produce a 20% salt solution? 50 x y - =
10% SALT SOLUTION
PURE WATER
20% SALT SOLUTION
Let x = amount of water evaporated
Let y = amount of new solution
First equation: x = y

8. Why is the percentage on the water 0%?50 x y - =
10% SALT SOLUTION
Pure water
20% SALT SOLUTION - =
% · amount
% · amount
% · amount
.10 · · x = · y
Why is the percentage on the water 0%?

9. 25 L of water must be evaporated.
Solve the system using the substitution method.
25 L of water must be evaporated.

10. Half-and-Half with 11% butterfat
Ex. 3: Milk with 3% butterfat was mixed with cream with 27% butterfat to produce 36 L of Half-and-Half with 11% butterfat content. How much of each was used? x y 36
Milk with 3% butterfat +
Cream with 27% butterfat =
Half-and-Half with 11% butterfat
Let x = amount of milk added
Let y = amount of cream added
First equation:
x + y = 36
This equation says we started with x liters of milk and are adding y liters of cream to produce 36 liters of Half-and-Half.

11. Half-and-Half with 11% butterfatx y 36
Milk with 3% butterfat
Cream with 27% butterfat
Half-and-Half with 11% butterfat + = + =
% · amount
% · amount
% · amount
.03·x ·y = .11·36
Solve the system:
24 L of milk and 12 L of cream are needed.

12. Ex. 4: A chemistry experiment calls for a 30% solution of copper sulfate. Mrs. Maiorca has 40 milliliters of 25% solution. How many milliliters of 60% solution should she add to make a 30% solution? 40 x y + =
25% solution
60% solution
30% solution + =
% · amount
% · amount
% · amount
Let x = amount 60% solution
Let y = amount of 30% solution
40 + x = y
.25(40) + .60(x) = .30y

13. Mrs. Maiorca needs to add 6.67 mL of the 60% solution.
40 + x = y
.25(40) + .60(x) = .30y
x = .30(40 + x)
x = x
.60x = x
.30x = 2
x ≈ 6.67
Mrs. Maiorca needs to add 6.67 mL of the 60% solution.