There are many situations where we desire to mix two or more substances together in precise ratio. These include mixing medicines in substances (like water) and mixing ingredients in a recipe. This section presents how to calculate the ratio of substances after mixing multiple solutions, and the reverse problem how to calculate the amount of each solution to mix for a desired ratio of substances.
In some circumstances we know the concentrations of substances in multiple solutions and how much of each has been combined. From this we can calculate the concentration of substances in the resulting solution.
Suppose we have a container with a solution of sugar water that is 22% sugar and another container with a solution of sugar water that is 16% sugar. If we combine 150 g of the first solution and 250 g of the second solution, what is the percent of sugar in the resulting solution? Round the percent to two decimal places.
To calculate a percent we need the amount of the part and the amount of the whole. We are told the total amount of each solution we are using, so we can calculate the whole (total amount of resulting solution) directly. Because they have the same units we can add.
We cannot simply add the amounts of sugar from each because we are told the percent rather than the amount. However, we know how to convert a percent and total amount to an amount of the part. We did this in Example 2.1.5.
Finally we can calculate the percent of sugar in the resulting mixture using the defintion of percent. The percent of sugar in the resulting mixture is
Suppose we have 11.3 lbs of 4140 steel which is a type of steel containing 40% carbon and 9.2 lbs of 4150 steel which contains 50% carbon. If we melt and mix these two metals, what is the resulting percent carbon? Round the final percent to two decimal places.
To calculate a percent we need the amount of the part and the total amount. We can calculate the total directly: \(T=11.3 \text{ lbs}+9.2 \text{ lbs}=20.5 \text{ lbs}\text{.}\) Note we know we can add these because both represented total amounts and they have the same units (g).
To calculate the part we need to know how much (rather than what percent) carbon was obtained from the two metals. We can calculate that from the given percents and amounts.
91% isopropyl alcohol is not a good agtent for sanitzing because it evaporates so quickly that bacteria are not exposed long enough to die. As a result we want to dilute it before use. Suppose we begin with 16.0 oz of a 91.0% isopropyl alcohol and add 4.00 oz of water to this mixture. What will the percent alcohol of the resulting solution be? These are measurements, so we will round using significant digits.
We know we can calculate the percent alcohol by dividing the amount of alcohol (part) by the total amount. We also know that we are not increasing the amount of alcohol (adding only water), so we can use the original percent alcohol and original amount to calculate the amount of alcohol.
If we added more water would the percent alcohol be greater or less? If we are using all of the alcohol solution, the amount of water we add determines the percent alcohol.
In the previous section we calculated the result of mixing two solutions. In this section the goal is to figure out how much water to add to achieve a specific concentration. That is, the previous section calculated in this section we solve.
If we start with 16.0 oz of 91.0% alcohol solution, how much water do we add to obtain (at least) 25.0 oz of a 55.0% alcohol solution? These are measurements, so we will round using significant digits.
The concentration (55.0%) is the primary goal here rather than the resulting amount (25.0 oz), because we need at least that amount: more is okay. As a result we will set up our calculation based on the desired percent alcohol.
This is a percent problem with the total alcohol unchanged and adding only some amount \(w\) of water. This matches the previous example, except that we are calculating the amount to add and know the resulting percent. However, we can write essentially the same equation. \(\frac{\text{part}}{\text{whole}} = \frac{A}{16.0+w}=0.550\) where \(A\) is the amount of alcohol. We do not use the 25.0 oz at this time. We will address that at the end.
Because we are adding only water, all of the alcohol comes from the initial solution. Thus we can calculate the amount of alcohol in the same way as before. \((0.910)16.0 \text{ oz} = 14.56 \text{ oz}\text{.}\) Thus we setup
We end up with \(16 \text{ oz}+10 \text{ oz} = 27 \text{ oz}\) of new solution. Because we have at least the 25 oz we need, and it is the correct concentration, we have achieved our goal. If we had added less water to get exactly 25 oz we would have had a more concentrated solution than desired.
Also note that the percent alcohol is \(\frac{0.910(16.0 \text{ oz})}{16.0 \text{ oz}+10 \text{ oz}}=0.56\) or 56%. This is not exactly 55% because of the rounding in the calculations. If we needed it to be closer to 55%, we would need to be able to measure more precisely.
