Skill Builder ES1

If 38.4 g of element A are needed to combine with 17.8 g of element B, then how many grams of element A are needed to combine with 52.3 g of element B?

Solution

We set up the proportions as follows:

Multiplying both sides of the equation by 52.3 g gives

Notice that grams cancel to leave us with an answer that is in the correct units. Always check to make sure that your answer has the correct units.

Skill Builder ES2

Solve to find the indicated variable.

1. $\frac{\text{16}\text{.4 g}}{\text{41}\text{.2 g}}=\frac{x}{18.\text{3 g}}$
2. $\frac{\text{2}\text{.65 m}}{\text{4}\text{.02 m}}=\frac{\text{3}\text{.28 m}}{y}$
3. $\frac{\text{3}\text{.27}\times {10}^{-3}\text{g}}{x}=\frac{\text{5}\text{.0}\times {10}^{-1}\text{g}}{3.2\text{g}}$
4. Solve for V₁: $\frac{P_1}{P_2}=\frac{V_2}{V_1}$
5. Solve for T₁: $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Solution

1. Multiply both sides of the equality by 18.3 g to remove this measurement from the denominator:

   $$\begin{array}{l}\text{(18}\text{.3 g)}\frac{\text{16}\text{.4}\bcancel{\text{g}}}{\text{41}\text{.2}\bcancel{\text{g}}}=\bcancel{\text{(18}\text{.3 g)}}\frac{x}{\bcancel{18.\text{3 g}}}\\ \text{7}\text{.28 g}=x\end{array}$$

2. Multiply both sides of the equality by 1/3.28 m, solve the left side of the equation, and then invert to solve for y:

   $$\begin{array}{c}\left(\frac{1}{3.28\text{m}}\right)\frac{\text{2}\text{.65 m}}{\text{4}\text{.02 m}}=\left(\frac{1}{\bcancel{3.\text{28 m}}}\right)\frac{\bcancel{3.\text{28 m}}}{y}=\frac{1}{y}\\ y=\frac{(4.02)(3.28)}{2.65}=4.98\text{m}\end{array}$$

3. Multiply both sides of the equality by 1/3.27 × 10⁻³ g, solve the right side of the equation, and then invert to find x:

   $$\begin{array}{c}\left(\frac{1}{\bcancel{\text{3}\text{.27}\times {10}^{-3}\text{g}}}\right)\frac{\bcancel{\text{3}\text{.27}\times {10}^{-3}\text{g}}}{x}=\left(\frac{1}{\text{3}\text{.27}\times {10}^{-3}\text{g}}\right)\frac{\text{5}\text{.0}\times {10}^{-1}\bcancel{\text{g}}}{\text{3}\text{.2}\bcancel{\text{g}}}=\frac{1}{x}\\ x=\frac{(3.2\text{g)(3}\text{.27}\times {10}^{-3}\text{g)}}{\text{5}\text{.0}\times {10}^{-1}\text{g}}=2.1\times {10}^{-2}\text{g}\end{array}$$

4. Multiply both sides of the equality by 1/V₂, and then invert both sides to obtain V₁:

   $$\begin{array}{l}\left(\frac{1}{V_2}\right)\frac{P_1}{P_2}=\left(\frac{1}{\bcancel{V_2}}\right)\frac{\bcancel{V_2}}{V_1}\\ \frac{P_2{V}_2}{P_1}=V_1\end{array}$$

5. Multiply both sides of the equality by 1/P₁V₁ and then invert both sides to obtain T₁:

   $$\begin{array}{l}\left(\frac{1}{\bcancel{P_1{V}_1}}\right)\frac{\bcancel{P_1{V}_1}}{T_1}=\left(\frac{1}{P_1{V}_1}\right)\frac{P_2{V}_2}{T_2}\\ \frac{1}{T_1}=\frac{P_2{V}_2}{T_2{P}_1{V}_1}\\ T_1=\frac{T_2{P}_1{V}_1}{P_2{V}_2}\end{array}$$

Skill Builder ES3

Convert each number to a percentage or a decimal.

1. 29.4%
2. 0.390
3. 101%
4. 1.023

Solution

1. $\frac{29.4}{100}\text{= 0}\text{.294}$
2. 0.390 × 100 = 39.0%
3. $\frac{101}{100}\text{= 1}\text{.01}$
4. 1.023 × 100 = 102.3%

Skill Builder ES4

Use percentages to answer the following questions, being sure to use the correct number of significant figures. Express your answer in scientific notation where appropriate.

1. What is the mass of hydrogen in 52.83 g of a compound that is 11.2% hydrogen?
2. What is the percentage of carbon in 28.4 g of a compound that contains 13.79 g of that element?
3. A compound that is 4.08% oxygen contains 194 mg of that element. What is the mass of the compound?

Solution

1. $\begin{array}{l}\text{52}\text{.83 g}\times \frac{\text{11}\text{.2}}{\text{100}}=\text{52}\text{.83 g}\times \text{0}\text{.112}=\text{5}\text{.92 g}\\ \end{array}$
2. $\frac{13.79\text{g carbon}}{\text{28}\text{.4 g}}\times 100=48.6\%\text{carbon}$

3. This problem can be solved by using a proportion:

   $$\begin{array}{lll}\frac{\text{4}\text{.08\% oxygen}}{\text{100\% compound}}\hfill & =\hfill & \frac{\text{194 mg}}{x\text{mg}}\\ x\hfill & =\hfill & \text{4}\text{.75}\times {\text{10}}^{\text{3}}\text{mg (or 4}\text{.75 g)}\end{array}$$

Skill Builder ES5

Use the information in Table 1.8 "Prefixes Used with SI Units" to convert each measurement. Be sure that your answers contain the correct number of significant figures and are expressed in scientific notation where appropriate.

1. 59.2 cm to decimeters
2. 3.7 × 10⁵ mg to kilograms
3. 270 mL to cubic decimeters
4. 2.04 × 10³ g to tons
5. 9.024 × 10¹⁰ s to years

Solution

1. $\text{59}\text{.2}\cancel{\text{cm}}\times \frac{\text{1}\cancel{\text{m}}}{\text{100}\cancel{\text{cm}}}\times \frac{\text{10 dm}}{\text{1}\cancel{\text{m}}}=\text{5}\text{.92 dm}$
2. $\text{3}\text{.7}\times {\text{10}}^{\text{5}}\cancel{\text{mg}}\times \frac{\text{1}\cancel{\text{g}}}{\text{1000}\cancel{\text{mg}}}\times \frac{\text{1 kg}}{\text{1000}\cancel{\text{g}}}=\text{3}\text{.7}\times {\text{10}}^{-\text{1}}\text{kg}$
3. $\text{270}\cancel{\text{mL}}\times \frac{\text{1}\cancel{\text{L}}}{\text{1000}\cancel{\text{mL}}}\times \frac{{\text{1 dm}}^{\text{3}}}{\text{1}\cancel{\text{L}}}=\text{270}\times {\text{10}}^{-3}{\text{dm}}^{\text{3}}=\text{2}\text{.70}\times {\text{10}}^{-\text{1}}{\text{dm}}^{\text{3}}$
4. $\text{2}\text{.04}\times {\text{10}}^{\text{3}}\cancel{\text{g}}\times \frac{\text{1}\cancel{\text{lb}}}{\text{453}\text{.6}\cancel{\text{g}}}\times \frac{\text{1 tn}}{\text{2000}\cancel{\text{lb}}}=\text{0}\text{.00225 tn}=\text{2}\text{.25}\times {\text{10}}^{-\text{3}}\text{tn}$
5. $\text{9}\text{.024}\times {\text{10}}^{\text{10}}\cancel{\text{s}}\times \frac{\text{1}\cancel{\text{min}}}{\text{60}\cancel{\text{s}}}\times \frac{\text{1}\cancel{\text{h}}}{\text{60}\cancel{\text{min}}}\times \frac{\text{1}\cancel{\text{d}}}{\text{24}\cancel{\text{h}}}\times \frac{\text{1 yr}}{\text{365}\cancel{\text{d}}}=\text{2}\text{.86}\times {\text{10}}^{\text{3}}\text{yr}$