Math 8H

Solving Chemical Mixture Problems

Algebra 1 Glencoe McGraw-Hill JoAnn Evans

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.

1 st ingredient

+

2 nd 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.

% · amount

+

% · amount

=

% · amount

Mr. Williams has 40 ml of a solution that is 50% acid. How much water should he add to make a solution that is 10% acid?

40

50% ACID SOLUTION

+

x PURE WATER

=

y 10% ACID SOLUTION

The problem asks for the amount of water Mr. Williams should add. We also need to know how much of the 10% solution he’ll end up with.

Let x = amount of water added Let y = amount of new solution

First equation: 40 + x = y

The first equation only addressed the amount of the liquids.

40 + x = y

The equation says that Mr. Williams 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).

% · amount

+

% · amount

=

% · amount

40 50% ACID SOLUTION

+

x PURE WATER

=

y 10% ACID SOLUTION

% · amount + % · amount = % · amount

.50 · 40 + .00 · x = .10 · y

Why is the percentage on the water 0%? Because the % tells what percentage of ACID is in each solution. There is no acid in the water added.

   .

40 50 (  x 40 )   y 0 x  .

10 y

Solve the system using the substitution method.

   .

40 50 (  x 40 )   y 0 x  .

10 y .

50 ( 40 )  0 x  .

10 ( 40  x )  20 4  0   4  4 .

10 x 16 160   .

10 x x

Mr. Williams needs to add 160 ml of water.

How many liters of acid should Mrs. Bartley add to 4 L of a 10% acid solution to make a solution that is 80% acid?

4 10% ACID SOLUTION

+

x PURE ACID

=

y 80% ACID SOLUTION

The problem asks for the amount of acid Mrs. Bartley should add. We also need to know how much of the new solution she’ll end up with.

Let x = amount of acid added Let y = amount of stronger acid solution

First equation: 4 + x = y

The first equation only addressed the amount of the liquids.

4 + x = y

weak acid + pure acid = stronger acid

The second equation in the system needs to address the strength of each solution (percentage of acid).

4 10% ACID SOLUTION

+

x PURE ACID

=

y 80% ACID SOLUTION

% · amount + % · amount = % · amount

.10 · 4 + 1.00 · x = .80 · y

Why is the percentage on the acid 100%? Because the % tells what percentage of ACID is in each solution. Pure acid is 100% acid.

   .

4  10 ( x 4 )   y 1 .

00 x  .

80 y

Solve the system using the substitution method.

   .

4  10 ( x 4 )   y 1 .

00 x  .

80 y .

10 ( 4 )  1 .

00 x  .

80 ( 4  x )  .

40 .

40  x   3 .

.

20 40  .

80 x  x  .

80 x 2 .

80   .

80 x .

80 x .

20 x x   2 .

80 14

Mrs. Bartley needs to add 14L of acid.

How many liters of water must Miss Elias EVAPORATE from 50 L of a 10% salt solution to produce a 20% salt solution?

50 10% SALT SOLUTION

-

x PURE WATER

=

y 20% SALT SOLUTION

Let x = amount of water lost Let y = amount of stronger salt solution

First equation: 50 - x = y

50 10% SALT SOLUTON

-

x Pure water

=

y 20% SALT SOLUTION

% · amount % · amount = % · amount

.10 · 50 - .00 · x = .20 · y

Why is the percentage on the water 0%? Because the % tells what percentage of SALT is in each solution. Pure water has 0% salt.

 50   .

10 (  x 50 )   y .

00 x  .

20 y

Solve the system using the substitution method.



50

 

.

10 (



x 50 )

 

y .

00 x



.

20 y

.

10 ( 50 )  .

00 x  .

20 ( 50  x ) 5  0  10  .

20 x  10 5  10  10  .

20 x  5   .

20 x 25  x

25 L of water must be evaporated.

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 Milk with 3% butterfat

+

y Cream with 27% butterfat

=

36 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.

x Milk with 3% butterfat

+

y Cream with 27% butterfat

= % · amount + % · amount

=

.03

· x + .27

· y = 36 Half and -Half with 11% butterfat

% · amount

.11

· 36

.

03 x  .

27 y .

03 ( 36  y 1 .

08  .

03 y )   .

27 y .

27 y  .

11 ( 36 ) Solve the system:    .

x  03 x y   .

36 27 y  .

11 ( 36 )

12 L of cream and 24 L of milk are needed.

y   .

11 ( 36 ) 3 1 .

08  .

.

24 y 24 y   3 .

96 2 .

88  12 .

96

A chemistry experiment calls for a 30% solution of copper sulfate. Mr. McGhee has 40 milliliters of 25% solution. How many milliliters of 60% solution should he add to make a 30% solution?

40 25% solution

% · amount + +

x 60% solution

% · amount = = % · amount Let x = amount 60% solution Let y = amount of 30% solution

y 30% solution

40 + x = y .25(40) + .60(x) = .30y

40 + x = y .25(40) + .60(x) = .30y

10 + .60x = .30(40 + x) 10 + .60x = 12 + .30x

.60x = 2 + .30x

.30x = 2 x ≈ 6.67

Mr. McGhee needs to add 6.67 milliliters of the 60% solution.