Adding a Pure Substance

Overview

Some mixture problems involve adding a substance that consists entirely of the component being tracked. Examples include adding pure salt to a salt solution, pure alcohol to an alcohol mixture, or pure metal to an alloy.

These problems may appear different from ordinary concentration problems, but mathematically they follow the same Mixture Principle:

The only new feature is that a pure substance has concentration

That means the component amount contributed by \(x\) units of pure substance is simply

No new formula is required.

Pure Means One Hundred Percent

Suppose \(x\) kilograms of pure copper are added to an alloy.

Because the material is pure copper, its copper concentration is

Using the Mixture Principle,

the amount of copper contributed is

Therefore,

Every unit added is also one full unit of the component being tracked.

The General Model

Suppose an existing mixture has concentration \(C\) and total quantity \(T\). Then its component amount is

Now suppose \(x\) units of pure component are added.

The pure substance contributes

units of the component.

The final mixture contains

units in total.

If the desired final concentration is \(C_f\), conservation gives

This equation is the standard model for adding a pure substance to an existing mixture.

Example 1: Adding Pure Salt

A tank contains \(10\) kilograms of a mixture that is \(20\%\) salt. How much pure salt must be added to make the mixture \(40\%\) salt?

Let \(x\) represent the kilograms of pure salt added.

The original mixture contains

kilograms of salt.

The pure salt contributes

kilograms of salt.

The final mixture has total mass

At \(40\%\) concentration, the final amount of salt is

Apply conservation:

Expand:

Subtract \(0.40x\):

Subtract \(2\):

Therefore,

So approximately

of pure salt must be added.

Checking the Result

The original mixture contained \(2\) kilograms of salt. Adding approximately \(3.33\) kilograms of pure salt gives

kilograms of salt.

The total mixture becomes

kilograms.

Then

So the final concentration is approximately

The result is consistent with the conditions of the problem.

Organizing the Problem in a Table

Reading across the component-amount column gives

The table helps show that a pure substance is not a special exception. It is simply a mixture row with concentration \(1.00\).

Why the Concentration Increases

When pure component is added, two things happen at the same time:

• The amount of the tracked component increases.
• The total amount of mixture also increases.

However, the added material is entirely made of the tracked component. Because its concentration is higher than the concentration of the original mixture, the final concentration rises.

This leads to a useful qualitative rule:

A pure substance is simply the most extreme case because its concentration is \(100\%\).

Example 2: Adding Pure Alcohol

A container holds \(15\) liters of a \(30\%\) alcohol solution. How many liters of pure alcohol must be added to produce a \(50\%\) solution?

Let \(x\) be the amount of pure alcohol added.

The original amount of alcohol is

Pure alcohol contributes

The final quantity is

The final alcohol amount is therefore

Conservation gives

Expand:

Subtract \(0.50x\):

Subtract \(4.5\):

Therefore,

So

must be added.

Solving the General Pure-Addition Equation

Begin with

Expand the right side:

Move the terms containing \(x\) to one side:

Factor:

Therefore,

This formula gives the amount of pure component required to raise an existing mixture from concentration \(C\) to concentration \(C_f\).

The ordinary conservation equation is usually easier to remember, but this derived form reveals how the required amount depends on the starting concentration, target concentration, and original quantity.

What the Formula Tells Us

Consider

Several relationships become visible.

If the starting amount \(T\) doubles while the concentrations stay the same, then \(x\) also doubles.

If the target concentration \(C_f\) moves farther above the starting concentration \(C\), more pure component is required.

As the target concentration approaches \(100\%\),

the denominator

approaches zero.

This means that obtaining a concentration arbitrarily close to \(100\%\) requires an increasingly large amount of pure substance.

Can an Impure Mixture Reach Exactly 100%?

Suppose a mixture initially contains some material other than the tracked component. Can adding a finite amount of pure component make the final concentration exactly \(100\%\)?

No.

The original non-component material remains in the mixture. Adding more pure component can make its proportion smaller and smaller, but it does not remove it.

The general equation helps show why. If

then

which requires division by zero.

There is no finite value of \(x\) that makes an initially impure mixture exactly pure merely by adding more pure material.

Approaching Purity

Although exactly \(100\%\) cannot be reached by finite addition, concentrations can approach \(100\%\).

Suppose \(10\) liters of a \(20\%\) solution are strengthened with pure component.

To reach \(80\%\):

To reach \(90\%\):

To reach \(99\%\):

As the target moves closer to \(100\%\), the required amount grows dramatically.

Example 3: Strengthening an Alloy

A \(24\)-kilogram alloy is \(25\%\) copper. How much pure copper should be added to make the resulting alloy \(40\%\) copper?

Let \(x\) be the kilograms of pure copper added.

The original amount of copper is

After adding pure copper, the amount of copper is

The final mass is

So

Expand:

Subtract \(0.40x\):

Subtract \(6\):

Therefore,

Thus,

must be added.

Reverse Problems: Finding the Final Concentration

Sometimes the amount of pure substance added is known and the final concentration must be determined.

Suppose \(12\) kilograms of a \(25\%\) mixture receive \(3\) kilograms of pure component.

The original component amount is

The pure substance adds another

So the final component amount is

The final total quantity is

Therefore,

The resulting mixture is

Example 4: Finding the Original Concentration

A \(10\)-liter solution has an unknown concentration. After \(5\) liters of pure component are added, the resulting \(15\)-liter mixture is \(60\%\). Find the original concentration.

Let \(C\) represent the original concentration.

The original component amount is

The pure component contributes

The final mixture contains

liters of the component.

Therefore,

Subtract \(5\):

Divide by \(10\):

So the original concentration was

Pure Substance as a Limit Case

In ordinary concentration problems, one source might have concentration \(C_2\). The conservation equation is

For a pure substance,

So the equation becomes

or simply

This demonstrates that pure-substance problems are not a separate mathematical species. They are ordinary concentration problems in which one concentration happens to equal \(1\).

Pure Addition Cannot Lower the Concentration

Suppose the existing concentration is \(30\%\). Adding material that is \(100\%\) component cannot produce a final concentration of \(20\%\).

The added substance is more concentrated than the original mixture, so the final concentration must rise.

In general, if

then the final concentration moves upward.

Therefore, when pure substance is added,

Method for Adding a Pure Substance

1. Identify the component being tracked. For example, salt, alcohol, copper, or another component.
2. Find the original component amount. Use \[ A=CT. \]
3. Represent the pure addition. If \(x\) units are added, the component amount added is also \[ x. \]
4. Determine the new total quantity. If \(x\) units are added to \(T\), the final total is \[ T+x. \]
5. Write the final component amount. Use \[ C_f(T+x). \]
6. Apply conservation. Write \[ CT+x=C_f(T+x). \]
7. Solve and check. Make sure the answer is positive and produces a final concentration between the starting concentration and \(100\%\).

Common Errors

• Treating pure substance as having concentration \(100\). Use \(1.00\), not \(100\), in the equation.
• Forgetting that the pure substance increases the total quantity. The final quantity is \(T+x\), not \(T\).
• Adding the concentration instead of the component amount. The pure contribution is \(x\), not \(1+x\) or \(100+x\).
• Assuming the final concentration can reach \(100\%\) with a finite addition. If the original mixture contains another component, that material remains present.
• Accepting a target below the starting concentration. Adding pure component raises concentration; it cannot dilute the mixture.
• Confusing component amount with total mixture amount. Keep the two quantities separate throughout the problem.

Practice Problems

1. A \(12\)-liter solution is \(25\%\) salt. How many liters of pure salt equivalent must be added to make the mixture \(40\%\) salt?
2. A \(20\)-kilogram alloy is \(30\%\) copper. How many kilograms of pure copper must be added to produce a \(50\%\) copper alloy?
3. A \(10\)-liter mixture is \(20\%\) alcohol. If \(5\) liters of pure alcohol are added, what is the final concentration?
4. A \(16\)-kilogram alloy is \(25\%\) gold. Pure gold is added until the alloy is \(40\%\) gold. How much pure gold is added?
5. A \(15\)-liter solution has an unknown concentration. After \(5\) liters of pure component are added, the final \(20\)-liter mixture is \(50\%\). Find the original concentration.
6. How much pure component must be added to \(8\) liters of a \(35\%\) mixture to obtain a \(65\%\) mixture?
7. Can pure component be added to a \(60\%\) mixture to produce a \(45\%\) mixture? Explain.
8. A \(30\)-kilogram mixture is \(40\%\) component. How much pure component must be added to make the final mixture \(70\%\)?

Practice Answers

1. \[ 0.25(12)+x=0.40(12+x) \] \[ 3+x=4.8+0.40x \] \[ 0.60x=1.8 \] \[ \boxed{x=3}. \]
2. \[ 0.30(20)+x=0.50(20+x) \] \[ 6+x=10+0.50x \] \[ 0.50x=4 \] \[ \boxed{x=8\text{ kg}}. \]
3. Original component: \[ 0.20(10)=2. \] After adding \(5\): \[ 2+5=7. \] Final quantity: \[ 10+5=15. \] Thus, \[ C_f=\frac{7}{15}\approx0.4667. \] So \[ \boxed{46.67\%}. \]
4. \[ 0.25(16)+x=0.40(16+x) \] \[ 4+x=6.4+0.40x \] \[ 0.60x=2.4 \] \[ \boxed{x=4\text{ kg}}. \]
5. Let \(C\) be the original concentration: \[ 15C+5=0.50(20). \] \[ 15C+5=10. \] \[ 15C=5. \] \[ C=\frac13. \] Therefore, \[ \boxed{33\tfrac13\%}. \]
6. \[ 0.35(8)+x=0.65(8+x) \] \[ 2.8+x=5.2+0.65x \] \[ 0.35x=2.4 \] \[ x=\frac{48}{7}\approx6.86. \] Therefore, \[ \boxed{6.86\text{ units approximately}}. \]
7. \[ \boxed{\text{No}}. \] Adding a \(100\%\) component to a \(60\%\) mixture can only increase the concentration above \(60\%\).
8. \[ 0.40(30)+x=0.70(30+x) \] \[ 12+x=21+0.70x \] \[ 0.30x=9 \] \[ \boxed{x=30\text{ kg}}. \]

Summary

A pure substance has concentration

Therefore, if \(x\) units of pure component are added, the amount of component contributed is

For an original mixture with concentration \(C\) and quantity \(T\), adding \(x\) units of pure component to reach final concentration \(C_f\) gives

Solving this equation for \(x\) gives

The deeper principle remains conservation. Pure-substance problems are simply ordinary mixture problems in which one source has concentration \(1\).

Continue the Exploration

Adding a pure component raises concentration. The next lesson examines the opposite process: lowering concentration by adding a substance that contains none of the component being tracked.

• Continue to: Dilution Problems

  Learn how adding a \(0\%\) substance increases the total quantity while leaving the amount of the tracked component unchanged.