2.9: Fractions and Percent
- Page ID
- 276353
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Fractions and percent essentially tell you the same thing, they describe the ratio of a part of the whole.
Fraction
\[Fraction=\frac{Part}{Whole}=\frac{Part}{\sum Parts}\; \\ \left( \sum \; \; \text{means sum of} \right )\]
Exercise \(\PageIndex{1}\)
Consider Mixing 23.5 g of table salt (NaCl) with 91 g of water
- What is the fraction salt?
- What is the fraction water?
- Answer a
-
\[ \text{fraction salt = } \frac{m_{salt}}{m_{salt}+m_{water}} = \frac{23.5g}{23.5g+91g} = \frac{23.5g}{114.5g}=0.20524=0.205 \nonumber \]
- Answer b
-
\[ \text{fraction water = } \frac{m_{water}}{m_{salt}+m_{water}} = \frac{91g}{23.5g+91g} = \frac{91g}{114.5g}=0.79476 =0.79 \]
Note, the number of significant figures in answer (a) above is different than answer (b).
The sum of the fractions equals 1, that is all the parts combine into the whole, and 1 is a defined number.
\[ \begin{align} fraction_{salt} \; +\; fraction_{water} &=\frac{m_{salt}}{m_{salt}+m_{water}}+\frac{m_{water}}{m_{salt}+m_{water}} \\ \nonumber \; \\ \nonumber &=\frac{m_{salt}+m_{water}}{m_{salt}+m_{water}} \nonumber \\ \; \nonumber \\ &=1 \end{align}\]
In general this can be expressed as the summation of each fraction starting with i=1, being summed up to i=n, when there are n parts (fractions)
\[\sum_{i=1}^{n}\left (fraction \right )_i=1 \]
So there was a second way to solve exercise 1b, if we know the answer to 1a
\[ \begin{align}fraction_{salt}+fraction_{water} & =1 \nonumber \\ \therefore \nonumber \\ fraction_{water} &=\; 1-fraction_{salt} \nonumber \\ &= \; 1-0.205 \nonumber \\ &= 0.795 \end{align}\]
Note
The number of significant digits in solving problems can vary depending on the mathematical technique you use
Percentages
Percentages are 100 times the fraction, and so we can redo all of the above material in terms of percents. Note, the 100 is a defined number and does not contribute to the significant figures (100% of a sample is defined as the entire sample). The symbol for percent is %.
\[Percent =Fraction(100)=\frac{Part}{Whole}(100)\]
Exercise \(\PageIndex{2}\)
Consider Mixing 23.5 g of table salt (NaCl) with 91 g of water
- What is the percent salt?
- What is the percent water?
- Answer a
-
\[ \% \; salt = \frac{m_{salt}}{m_{salt}+m_{water}}(100) = \frac{23.5g}{23.5g+91g}(100) = \frac{23.5g}{114.5g}(100)=20.524 \%=20.5 \% \]
- Answer b
-
\[ \% \; water = \frac{m_{water}}{m_{salt}+m_{water}}(100) = \frac{91g}{23.5g+91g}(100) = \frac{91g}{114.5g}(100)=79.476 \% =79 \% \]
Note, the number of significant figures in answer (a) above is different than answer (b).
The sum of the percentages equals 100, that is all the parts combine into 100%.
\[ \begin{align} \%_{salt} \; +\; \%_{water} &=\frac{m_{salt}}{m_{salt}+m_{water}}(100)+\frac{m_{water}}{m_{salt}+m_{water}}(100)\\ \nonumber \; \\ \nonumber &=\frac{m_{salt}+m_{water}}{m_{salt}+m_{water}}(100) \nonumber \\ \; \nonumber \\ &=1(100)=100 \%\end{align}\]
In general this can be expressed as the summation of each fraction starting with i=1, being summed up to i=n, when there are n parts (fractions)
\[\sum_{i=1}^{n}\left (percentage \right )_i=100 \% \]
So there was a second way to solve exercise 1b, if we know the answer to 1a
\[ \begin{align}\%_{salt}+\%_{water} & =100 \% \nonumber \\ \therefore \nonumber \\ \% \_{water} &=\; 100 \% - \%_{salt} \nonumber \\ &= \; 100 \% -20.5 \% \nonumber \\ &= 79.5 \% \end{align}\]
Exercise \(\PageIndex{3}\)
What mass of Water do you need to add to 55.4g of salt to make a solution that is 15.0% Salt?
Hint: Solve this in terms of fractions, where fraction of salt = \(\frac{\%salt}{100}\)
- Answer
-
fs=fraction salt, ms=mass salt, mw=mass water, if percent salt = 15.0%, the fs= 0.150
\[f_{s}=\frac{m_{s}}{m_{s}+m_{w}} \\ \; \\ \text{solve for }m_w \\ \; \\
f_{s}\left(m_{s}+m_{w}\right)=m_{s} \\ \; \\
f_{s} m_{s}+f_{s} m_{w}=m_{s} \\ \; \\
f_{s} m_{w}=m_{s}-f_{s} m_{s}=m_{s}\left(1-f_{s}\right) \\ \; \\
m_{w}=\frac{\left(1-f_{s}\right)}{f_{s}} m_{s}=\frac{(1-.150)}{.150} 55.4 g=\frac{850}{.150} 55.4 g=314 g
\]
Fractions & Percent are Unitless
Lets look at a bag with of 100 nickels and 100 dimes, and measure the mass on two different scales and then calculate the percent of the mass in the bag that is dues to the nickles and the mass that is dimes.
| Mass in grams | mass in ounces | |
| 100 nickels | 500.0 g | 17.63 oz |
| 100 dimes | 226.8 g | 8.000 oz |
| total mass | 726.8 g | 25.63 |
In units of grams
\[\% nickle = \frac{500g}{726.8g} (100)= 68.79 \% \;\;\; \% dimes=100 \% - 68.79%=31.21\% \]
In units of ounces
\[\% nickle = \frac{17.63 oz}{25.63 oz} (100)= 68.79 \% \;\;\; \% dimes=100 \% - 68.79%=31.21\% \]
Note, both grams and ounces are units are measurements of mass, and the mass percent is independent of the unit. But if you are measuring the number of entities, the above bag is 50% nickels and 50% dimes.
Test Yourself
Homework: Section 2.9
Interactive Quiz \(\PageIndex{1}\)
Solutions to the above problems can be found in the homework section 2.9.