1.5 Mathematical Treatment of Measurement Results

Conversion factors have been discussed for metric-to-metric conversions earlier in the chapter, but conversions between English and metric units should also be discussed. Dimensional analysis, the factor label method, is where units of quantities are subjected to the same mathematical operations as their associated numbers.

Remember that a conversion factor is a fraction in which the same quantity is expressed in one way in the numerator and another way in the denominator, giving a ratio equal to 1. The previous example discussed was the conversion factor between minutes and hours.

\[\dfrac{60~\mathrm{min}}{1~\mathrm{h}} ~~\mathrm{or}~~ \dfrac{1~\mathrm{h}}{60~\mathrm{min}}\] It is clear that 60 does not equal one, however, when units are included 60 minutes does equal to one hour. These are two values that are equal. But you know if we were looking at them numerically, they would not look like they were representing the same thing. This can be a confusing concept at first, but it is still important to note.

Common English to metric equality statements can be found in the following table.

Length	Volume	Mass
1 m = 1.0936 yd	1 L = 1.0567 qt	1 kg = 2.2046 lb
1 in. = 2.54 cm (exact)	1 qt = 0.94635 L	1 lb = 453.59 g
1 km = 0.62137 mi	1 ft³ = 28.317 L	1 (avoirdupois) oz = 28.349 g
1 mi = 1609.3 m	1 tbsp = 14.787 mL	1 (troy) oz = 31.103 g

The English to metric conversion between inches and centimeters is

\[1~\mathrm{in} = 2.54~\mathrm{cm}\]

This equality statement, like the metric-to-metric conversion equality statements can be used to create the conversion factor between inches and centimeters.

\[\dfrac{1~\mathrm{in}}{2.54~\mathrm{cm}} ~~\mathrm{or}~~ \dfrac{2.54~\mathrm{cm}}{1~\mathrm{in}}\] Again, since the numerator is equal to the denominator, this is a ratio that is equal to 1, or a conversion factor.

Conversion factors, as we have seen previously, are useful in that they allow the conversion from one unit to another. Notice that in the example conversion factor above, the units of inches and centimeters are both present. This is an indicator of what two units this conversion factor will change the units between.

Using conversion factors in problems is called dimensional analysis or the factor label method. This type of mathematical operation allows the conversion to move from one unit to another.

Example:

Convert 12.00 inches to meters.

First, as done previously with the metric conversion, start by writing down what’s given in the problem. In this example problem, that is twelve inches.

\[\left ( \dfrac{12.00~\mathrm{in}}{} \right )\]

Recall that a conversion factor allows the movement from one unit to another. The next step requires identifying a conversion factor that allows movement from inches to meters (shown below). This relationship is given in the table above where the equality statement between inches and centimeters is given. Centimeters is not meters, but centimeters be converted from centimeters to meters using the prefix chart.

\[1~\mathrm{in} = 2.54~\mathrm{cm}\]

Now it is necessary to identify if inches or centimeters should be put on the top of our conversion factor. Remember to use the one that allows for the canceling of units.

\[\dfrac{1~\mathrm{in}}{2.54~\mathrm{cm}} ~~\mathrm{or}~~ \dfrac{2.54~\mathrm{cm}}{1~\mathrm{in}}\]

In order to move from inches to centimeters, the inches must cancelIn order to cancel the inches, the 2.54 cm must be on top and the one inch on bottom. From this point, the conversion can be accomplished, with a final step conversion from centimeter to meter.

\[\left ( \dfrac{12.00~\mathrm{in}}{} \right ) \left ( \dfrac{2.54~\mathrm{cm}}{1~\mathrm{in}} \right ) \left ( \dfrac{1~\mathrm{m}}{100~\mathrm{cm}} \right ) = 0.3048~\mathrm{m}\]

Now that the math is completed, the final answer should be put into the correct number of significant figures. The measurement values in the math and the type of math should be used to identify the correct number of significant figures in the final answer. Since the math is multiplication/division, the measurement with the smallest number of significant figures will determine the number of significant figures in the final answer. Note that the metric conversions have infinite significant figures, the inch is the value set for the centimeter measurements, and lastly the 12.00 inches is also a measurement with 4 significant figures. The smallest number of significant figures in the math is the 2.54 cm, therefore the final answer should have 3 significant figures. The final answer should then be reported as 0.304 meters.

Practice – Conversions

Convert 20.0 milligrams of gold into pounds.

Solution

\[\begin{align*} \left ( \dfrac{20.0~\mathrm{mg}}{} \right ) \left ( \dfrac{1~\mathrm{g}}{10^3~\mathrm{mg}} \right ) \left ( \dfrac{1~\mathrm{lb}}{453.6~\mathrm{g}} \right ) = 4.41\times 10^{-5}~\mathrm{lb} \end{align*}\]

Practice – Dimensional Analysis

The store is sold out of Tide washing detergent. You pick up a bottle called Generic Liquid Washing Soda Soap. It says it is made of sodium carbonate. You know from the label that there are 5.1864×10²⁴ atoms of sodium in each measuring cup. The measuring cup holds enough for a large load of laundry, 200.0 mL of soap. You are only doing a medium load of laundry, so only 100.0 mL of soap is required. How many grams of sodium carbonate will be present in the wash?

Solution

Carry out the following conversions:

mL → cup
cup → Na atoms
Na atoms → mol Na
mol Na → mol Na₂CO₃
mol Na₂CO₃ → g Na₂CO₃

\[\begin{align*} \left ( \dfrac{100.0~\mathrm{mL}}{~\mathrm{}} \right ) \left ( \dfrac{1~\mathrm{cup}}{200.0~\mathrm{mL}} \right ) \left ( \dfrac{5.1864\times 10^{24}~\mathrm{Na}}{1~\mathrm{cup}} \right ) \left ( \dfrac{1~\mathrm{mol~Na}}{6.022\times 10^{23}~\mathrm{Na}} \right ) \left ( \dfrac{1~\mathrm{mol~Na_2CO_3}}{2~\mathrm{mol~Na}} \right ) \left ( \dfrac{105.99~\mathrm{g}}{1~\mathrm{mol}} \right ) = 228.2~\mathrm{g} \end{align*}\]

Practice – Dimensional Analysis

You are farmer Joe and you just bought a field in the country. Your field is 16.0 acres (1 acre = 43560 ft²) . Your cow needs to have grass for food, so you decide that you need to fertilize your field. You chose to go with ammonium nitrate for your fertilizer due to its low cost. Ammonium nitrate only cost $0.02 per 7.523×10²² molecules. If your field requires that 5.00 grams of ammonium nitrate be spread every square foot, how many grams of ammonium nitrate will be required to fertilize your field? How much (in $) will it cost to fertilize the field?

Solution

Part A

Carry out the following conversions:

acres → ft²
ft² → g

\[\begin{align*} \left ( \dfrac{16.0~\mathrm{acres}}{~\mathrm{}} \right ) \left ( \dfrac{43560~\mathrm{ft^2}}{1~\mathrm{acre}} \right ) \left ( \dfrac{5.00~\mathrm{g~NH_4NO3}}{1~\mathrm{ft^2}} \right ) &= 3484800~\mathrm{g~NH_4NO_3} \\[2ex] &\approx 3.48\times 10^{6}~\mathrm{g~NH_4NO_3} \end{align*}\]

Part B

Carry out the following conversions:

g → mol
mol ² → particles
particles → cents
cents → dollars

Be sure to use the unrounded result from Part A.

\[\begin{align*} \left ( \dfrac{3484800~\mathrm{g~NH_4NO_3}}{~\mathrm{}} \right ) \left ( \dfrac{1~\mathrm{mol}}{80.043~\mathrm{g}} \right ) \left ( \dfrac{6.022\times 10^{23}~\mathrm{particles}}{1~\mathrm{mol}} \right ) \left ( \dfrac{2~\mathrm{cents}}{7.523\times 10^{22}~\mathrm{particles}} \right ) \left ( \dfrac{\$1.00}{100~\mathrm{cents}} \right ) = \$6,970 \end{align*}\]

Practice – Dimensional Analysis

In a regular coke, there are 10.6 g of sugar (C₁₂H₂₂O₁₁) in every 100 mL. A coke isn’t only 100 mL, in fact there are 591 mL in a typically purchased coke bottle. Johnny loves to play video games and drink coke. Johnny drank 1.689×10²⁵ molecules of sugar while binge drinking coke at the last gaming convention. There are 4.00 calories in every 1 gram of sugar and 3500.0 calories in every 1.00 lb.

How many calories did Johnny consume?
What is his new weight after the gaming convention if he previously weight 284.00 lb and had no other contribution to weight gain or loss?
How many cokes did Johnny drink?

Solution

Part A

Carry out the following conversions:

particles of sugar → mol
mol → g
g → cal

\[\begin{align*} \left ( \dfrac{1.689\times 10^{25}~\mathrm{particles}}{~\mathrm{}} \right ) \left ( \dfrac{1~\mathrm{mol}}{6.022\times 10^{23}~\mathrm{particles}} \right ) \left ( \dfrac{342.3~\mathrm{g}}{1~\mathrm{mol}} \right ) \left ( \dfrac{4.00~\mathrm{cal}}{1~\mathrm{g}} \right ) &= 38402.172~\mathrm{cal} \\[2ex] &\approx 3.84\times 10^{4}~\mathrm{cal} \end{align*}\]

Part B

Carry out the following conversion:

cal → lb.

Be sure to use the unrounded result from Part A.

\[\begin{align*} \left ( \dfrac{38402.172~\mathrm{cal}}{~\mathrm{}} \right ) \left ( \dfrac{1~\mathrm{lb}}{3500.0~\mathrm{cal}} \right ) = 10.97~\mathrm{lb} \end{align*}\]

Determine Johnny’s new weight.

\[\begin{align*} 284.00~\mathrm{lb} + 10.97~\mathrm{lb} = 294.97~\mathrm{lb} \end{align*}\]

Part C

Carry out the following conversions:

particles of sugar → mol
mol → g
g → 100.0 mL
100 mL → cans of coke

\[\begin{align*} \left ( \dfrac{1.689\times 10^{25}~\mathrm{particles}}{~\mathrm{}} \right ) \left ( \dfrac{1~\mathrm{mol}}{6.022\times 10^{23}~\mathrm{particles}} \right ) \left ( \dfrac{342.3~\mathrm{g~sugar}}{1~\mathrm{mol}} \right ) \left ( \dfrac{100.0~\mathrm{mL}}{10.6~\mathrm{g~sugar}} \right ) \left ( \dfrac{1~\mathrm{coke}}{591~\mathrm{mL}} \right ) &= 153.25~\mathrm{cokes} \\[2ex] &\approx 153~\mathrm{cokes} \end{align*}\]

Practice – Conversions

Your swimming pool holds 13,500 gallons. Unfortunately there has been a drought, and it only has 46,992 quarts. You decide it is time to fill the swimming pool up and treat the pool water with chlorine.

If your hose adds water at a rate of 757.1 mL s^–1, how many hours will it take to fill your swimming pool to its maximum capacity? 4 qt = 1 gal
How many chlorine tablets are required for your refilled pool if you use 2 chlorine tablets per 10,000 gallons?

Solution

Part A

Determine how much water is needed to fill the pool up.

\[\begin{align*} V_{\mathrm{remaining}} &= V_{\mathrm{capacity}} - V_{\mathrm{present}} \\ &= 13,500~\mathrm{gal} - \left [ \left ( \dfrac{46,992~\mathrm{qt}}{} \right ) \left ( \dfrac{\mathrm{gal}}{4~\mathrm{qt}} \right ) \right ] \\ &= 1,752~\mathrm{gal} \end{align*}\]

Convert V_remaining to mL.

\[\begin{align*} V_{\mathrm{remaining}} &= \left ( \dfrac{1,752~\mathrm{gal}}{} \right ) \left ( \dfrac{3.785412~\mathrm{L}}{\mathrm{gal}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[2ex] &= 6.632\times 10^{6}~\mathrm{mL} \end{align*}\]

Determine how many hours it takes to fill the pool.

\[\begin{align*} t_{\mathrm{fill}} &= \dfrac{V_{\mathrm{remaining}}}{\mathrm{rate}} \\[2ex] &= \left ( \dfrac{6.632\times 10^{6}~\mathrm{mL}}{757.1~\mathrm{mL~s^{-1}}} \right ) \left ( \dfrac{\mathrm{min}}{60~\mathrm{s}} \right ) \left ( \dfrac{\mathrm{h}}{60~\mathrm{min}} \right ) \\[2ex] &= 2.43~\mathrm{h} \end{align*}\]

Part B

Determine the number of tablets needed to treat the water.

\[\begin{align*} \left ( \dfrac{13,500~\mathrm{gal}}{} \right ) \left ( \dfrac{2~\mathrm{tablets}}{10,000~\mathrm{gal}} \right ) = 2.7~\mathrm{tablets} \end{align*}\]