Math 8H

Algebra 1 Glencoe McGraw-Hill JoAnn Evans

Solving

Chemical Mixture Problems

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.

+ =cost · amount

cost · amount cost · amount

1st ingredie

nt

2nd ingredient

mixture

+ =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?

+ =50% ACID SOLUTION

PURE WATE

R

10% ACID SOLUTION

The problem asks for the amount of water Mr. Williams should add.

Let x = amount of water added

We also need to know how much of the 10% solution he’ll end up with.

Let y = amount of new solution

First equation: 40 + x = y

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+ =

+ =50% ACID SOLUTION

PURE WATER

10% ACID SOLUTION

% · amount

% · amount =+ % · amount

.50 · 40 + .00 · x = .10 · y

y10.x0)40(50.

yx40

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 x y

y10.x0)40(50.

yx40Solve the system using the substitution method.

)x40(10.x0)40(50.

x10.4020 44

x10.16

x160

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?

+ =10% ACID SOLUTION

PURE ACID

80% ACID SOLUTION

The problem asks for the amount of acid Mrs. Bartley should add.

Let x = amount of acid added

We also need to know how much of the new solution she’ll end up with.

Let y = amount of stronger acid solution

First equation: 4 + x = y

4 x y

4 + x = y

The first equation only addressed the amount of the liquids.

weak acid + pure acid = stronger acid

The second equation in the system needs to address the strength of

each solution (percentage of acid).

+ =10% ACID SOLUTION

PURE ACID

80% ACID SOLUTION

% · amount % · amount=+ % · amount

.10 · 4 + 1.00 · x = .80 · y

y80.x00.1)4(10.

yx4Why is the

percentage on the acid 100%?

Because the % tells what

percentage of ACID is in each

solution. Pure acid is 100% acid.

4 x y

)x4(80.x00.1)4(10.

x80.20.3x40. 40.40.

x80.80.2x x80.x80.

80.2x20. 14x

y80.x00.1)4(10.

yx4Solve the system using the substitution method.

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?

- =10% SALT SOLUTION

PURE WATER

20% SALT SOLUTION

Let x = amount of water lost

Let y = amount of stronger salt solution

First equation: 50 - x = y

50 x y

- =10% SALT SOLUTON

Pure water

20% SALT SOLUTION

% · amount

% · amount

=- % · amount

.10 · 50 - .00 · x = .20 · y

y20.x00.)50(10.

yx50Why is the percentage

on the water 0%? Because the % tells what percentage of

SALT is in each solution. Pure water has 0% salt.

50 x y

Solve the system using the substitution method.

y20.x00.)50(10.

yx50

)x50(20.x00.)50(10.

x20.1005

x20.105

1010

x20.5

x25

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?

+ =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.

x y 36

+ =Milk with

3% butterfat

Cream with 27% butterfat

Half-and-Half with 11% butterfat

% · amount

% · amount

=+ % · amount

.03·x + .27·y = .11·36

)36(11.y27.x03.

36yx

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

12y

88.2y24.

96.3y24.08.1

96.3y27.y03.08.1

)36(11.y27.)y36(03.

)36(11.y27.x03.

Solve the system:

x y 36

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?

+ =25% solution

60% solution

30% solution

% · amount% · amount % · amount+ =Let x = amount 60% solution

Let y = amount of 30% solution40 + x = y

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

40 x y

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.