1B: Review of the Tools of Quantitative Chemistry

 

SI Units

Exercise \(\PageIndex{1}\)

The SI base unit of length is the ____.

1. newton
2. liter
3. meter
4. candela
5. Ångstrom unit

Answer

3. meter (see 1B.3.1: SI Base Units)

Exercise \(\PageIndex{2}\)

The SI base unit of temperature is the ____.

1. Kelvin
2. calorie
3. Fahrenheit
4. Ångstrom
5. kilocalorie

Answer

1. Kelvin (see 1B.3.1: SI Base Units)

Exercise \(\PageIndex{3}\)

Which of the following is not a base SI unit?

1. meter
2. kelvin
3. calorie
4. second
5. mole

Answer

3. calorie (see 1B.3.1: SI Base Units)

Exercise \(\PageIndex{4}\)

The absolute zero point on the Kelvin scale is equal to _____.

1. –273.15 °C
2. 0 °C
3. 14.5 °C
4. 25 °C
5. –373.15 ⁰C

Answer

1. –273.15 °C (see 1B.5.2 Temperature Conversions)

 

SI Prefixes

Exercise \(\PageIndex{5}\)

Give the name of the prefix and the quantity indicated by the following symbols that are used with SI base units.

1. c
2. d
3. G
4. k

Answer a

centi-, × 10⁻²

Answer b

deci-, × 10⁻¹

Answer c

Giga-, × 10⁹

Answer d

kilo-, × 10³

Exercise \(\PageIndex{6}\)

Give the name of the prefix and the quantity indicated by the following symbols that are used with SI base units.

1. m
2. n
3. p
4. T

Answer a

milli-, × 10⁻³

Answer b

nano-, × 10⁻⁹

Answer c

pico-, × 10⁻¹²

Answer d

tera-, × 10¹²

 

Experimental Error

Exercise \(\PageIndex{7}\)

The following archery targets show marks that represent the results of four sets of measurements. Which target shows

1. a precise but inaccurate set of measurements?
2. an accurate but imprecise set of measurements?
3. a set of measurements that is both precise and accurate?
4. a set of measurements that is neither precise nor accurate?

Answer

1. B 
2. A
3. C
4. D (see 1B.2:1.1 Accuracy, 1B.2:1.2 Precision)

Exercise \(\PageIndex{8}\)

Classify the following sets of measurements as accurate, precise, both, or neither.

1. Checking for consistency in the weight of chocolate chip cookies: 17.27 g, 13.05 g, 19.46 g, 16.92 g
2. Testing the volume of a batch of 25-mL pipettes: 27.02 mL, 26.99 mL, 26.97 mL, 27.01 mL
3. Determining the purity of gold: 99.9999%, 99.9998%, 99.9998%, 99.9999% 

Answer

a. neither

b.precise

c.both (see 1B.2:1.1 Accuracy, 1B.2:1.2 Precision)

Significant Figures

Exercise \(\PageIndex{9}\)

Give the number of significant figures in each measured value.

1. 5.87
2. 0.031
3. 52.90
4. 0.2001
5. 500

Answer a

3 (see 1.B2:2.1 Significant Figure Rules)

Answer b

2 (see 1.B2:2.1 Significant Figure Rules)

Answer c

4 (see 1.B2:2.1 Significant Figure Rules)

Answer d

4 (see 1.B2:2.1 Significant Figure Rules)

Answer e

1 (see 1.B2:2.1 Significant Figure Rules),  note, some texts will say it is not determined.  That is, you do not know. We will use the convention that trailing zeros are not significant if there is not a decimal, but if your instructor uses a different convention, play the game and use the rules your class uses.

Exercise \(\PageIndex{10}\)

Give the number of significant figures in each measured value.

1. 0.0053
2. 2300
3. 32.00
4. 34.483
5. 3450000

Answer a

2 (see 1.B2:2.1 Significant Figure Rules)

Answer b

2 (see 1.B2:2.1 Significant Figure Rules)

Answer c

4 (see 1.B2:2.1 Significant Figure Rules)

Answer d

5 (see 1.B2:2.1 Significant Figure Rules)

Answer e

3 (see 1.B2:2.1 Significant Figure Rules)

 

SI Prefix Conversion

Exercise \(\PageIndex{1}\)

Convert the following:

1. 32.4 ng to _____ Mg
2. 56.4 T bytes to _____ M bytes

Answer a

3.24 x 10^-14 Mg

\[ 32.4ng \frac{10^{-6}Mg}{10^9ng} = 3.24*10^{-14}Mg \\ \text{or the long way} \\ (32.4ng)(\frac{10^{-9}g}{1 ng})(\frac{1Mg}{10^{6}g})=32.4*10^{(-9-6)}Mg=32.4*10^{-15}Mg= 3.24*10^{-14}Mg \]

Note:  You should be able to do this the first way without writing everything out (you add exponents on the numerator and subtract on the denominator).  see section 1B.3.5

Answer b

5.64 x 10⁷ M bytes

\[(56.4Tbytes)(\frac{10^{12}bytes}{1 Tbytes})(\frac{1Mbytes}{10^{6}bytes})=56.4*10^{(12-6)}Mbytes=56.4*10^{6}Mbytes= 5.64*10^{7}Mbytes\]

Exercise \(\PageIndex{2}\)

Convert the following:

1. 34.6 n sec to _____ G sec
2. 4.0 Mg to _____ kg

Answer a

3.46 x 10^-17 G sec

Answer b

4.0 x 10³ kg

Exercise \(\PageIndex{3}\)

Convert the following:

1. 3.57 x 10¹⁴ µL to _____ mL
2. 4.337 x 10¹⁴ TL to _____ µL

Answer a

3.57 x 10¹¹ mL

\[(3.57*10^{14}\mu L)(\frac{10^{3}mL}{10^{6}\mu L})=3.57*10^{\left ( 14+3-6 \right )}\mu L=3.57*10^{11}\mu L\]

Answer b

4.337 x 10³² µL

\[(4.337*10^{14} TL)(\frac{10^{6}\mu L}{10^{-12} TL})=4.337*10^{\left ( 14+6+12 \right )}\mu L=4.337*10^{32}\mu L\]

Exercise \(\PageIndex{4}\)

Convert the following:

1. 4.78 x 10²⁴ fg to _____ Mg
2. 2.00 x 10^-4 µsec to _____ psec

Answer a

4.78 x 10³ Mg

Answer b

2.00 x 10² psec

Conversions Between Different Scales

Exercise \(\PageIndex{5}\)

Convert the following:

1. 1.00 mL to _____ in³ (1 mL = 1 cm³)
2. 1.00 TL to _____ ft³ (1 mL = 1 cm³)

Answer a

0.0610 in³

\[(1.00 mL)(\frac{1 cm^{3}}{1 mL})(\frac{1 in}{2.54 cm})^{3}=0.0610 in^{3}\]

Answer b

3.53 x 10¹⁰ ft³

\[(4.337*10^{14} TL)(\frac{10^{6}\mu L}{10^{-12} TL})=4.337*10^{\left ( 14+6+12 \right )}\mu L=4.337*10^{32}\mu L\]

 

Converting Floating Decimal Point Numbers to Scientific Notation

Exercise \(\PageIndex{6}\)

Convert the following to scientific notation:

1. 0.00045602
2. 50444000
3. 253

Answer a

4.5602 x 10^-4

Answer b

5.0444000 x 10⁷

Answer c

2.53 x 10²

Exercise \(\PageIndex{7}\)

Convert the following to scientific notation:

1. 50444000
2. 32.010
3. 0.0034

Answer a

5.0444 x 10⁷

Answer b

3.2010 x 10¹

Answer c

3.4 x 10^-3

 

Mathematical Operations with Scientific Notation and Significant Figures

Exercise \(\PageIndex{8}\)

Solve the following problems:

1. \begin{equation}
   3.03860 \times 10^{32}+3.7 \times 10^{30}+3.68 \times 10^{31}
   \end{equation}
2. \begin{equation}
   3.03860 \times 10^{-32}+3.7 \times 10^{-30}+3.68 \times 10^{-31}
   \end{equation}

Answer a

\( \begin{align}\nonumber & 3.03860 \times 10^{32}\\\nonumber &0.037 \times 10^{32}\\\nonumber+& 0.368\times 10^{32}\\ \hline & 3.4436 \times 10^{32} \end{align}\)

The answer is \(3.444\times 10^{32}\) because two of the least precise value was known to the thousandths (section 1B3.4.2)

Answer b

\( \begin{align}\nonumber & 3.7 \times 10^{-30}\\\nonumber &0.368 \times 10^{-30}\\\nonumber+& 0.0303860\times 10^{-30}\\ \hline & 4.098386\times 10^{-30} \end{align}\)

The answer is \(4.1\times 10^{-30}\) because two of the least precise value was known to the thousandths (section 1B3.4.2)

Exercise \(\PageIndex{9}\)

Solve the following problems:

1. \begin{equation}
   \frac{\left(9.7 \times 10^{583}\right)\left(7.339 \times 10^{98}\right)}{\left(5.432 \times 10^{-645}\right)\left(3.446 \times 10^{-4484}\right)}
   \end{equation}
2. \begin{equation}
   \frac{\left(9.7 \times 10^{583}\right)\left(7.339 \times 10^{-98}\right)}{\left(5.432 \times 10^{-645}\right)\left(3.446 \times 10^{4484}\right)}
   \end{equation}

Answer a

\(\frac{(9.7\times10^{583})(7.339\times10^{98})}{(5.432\times10^{-645})(3.446\times10^{-4484})}\\(\frac{9.7\times7.339}{5.432\times3.446})(10^{583+98-(-645)-(-4484)})\\3.80306\times10^{5810}\)

The answer is \(3.8\times 10^{5810}\) (section 1B3.4.2)

Answer b

\(\frac{(9.7\times10^{583})(7.339\times10^{-98})}{(5.432\times10^{-645})(3.446\times10^{4484})}\\(\frac{9.7\times7.339}{5.432\times3.446})(10^{583-98-(-645)-(-4484)})\\3.80306\times10^{5614}\)

The answer is \(3.8\times 10^{5614}\)  (section 1B3.4.2)

Exercise \(\PageIndex{10}\)

Solve the following problems:

1. \begin{equation}
   \frac{\left(9.7 \times 10^{-583}\right)\left(7.339 \times 10^{-98}\right)}{\left(5.432 \times 10^{645}\right)\left(3.446 \times 10^{4484}\right)}
   \end{equation}
2. \begin{equation}
   \frac{\left(9.70 \times 10^{5}+6.3 \times 10^{4}\right)\left(3.33 \times 10^{5}\right)}{\left(5.00 \times 10^{45}\right)\left(3.20 \times 10^{-5}\right)}
   \end{equation}

Answer a

\(\frac{(9.7\times10^{-583})(6.3\times10^{-98})}{(5.432\times10^{645})(3.446\times10^{4484})}\\(\frac{9.7\times7.339}{5.432\times3.446})(10^{-583-98-645-4484})\\3.80306\times10^{-5810}\)

The answer is \(3.8\times 10^{-5810}\) (section 1B3.4.2)

Answer b

\(\frac{(9.70\times10^5\;+\;6.30\times10^4)(3.33\times10^5)}{(5.00\times10^{45})(3.20\times10^{-5})}\\\frac{(9.70\times10^5\;+\;0.63\times10^5)(3.33\times10^5)}{(5.00\times10^{45})(3.20\times10^{-5})}\\\frac{(10.33\times10^5)(3.33\times10^5)}{(5.00\times10^{45})(3.20\times10^{-5})}\\(\frac{10.33\times3.33}{5.00\times3.20})(10^{5+5-45-(-5)})\\2.1499\times10^{-30}\)

The answer is \(2.15\times 10^{-30}\) (section 1B3.4.2)

 

Conversions

Exercise \(\PageIndex{1}\)

To prepare for a laboratory period, a student lab assistant needs 125 g of a compound. A bottle containing 1/4 lb is available. Did the student have enough of the compound?

Answer

\(1 \;lb\;=\;454\;g\\( \frac{1\;\textcolor{red}{ \cancel{lb}}}{4}) ( \frac{454 \; g}{1 \;\textcolor{red}{ \cancel{lb}}})=113\;g\)

No, 1/4 lb is only about 113 grams. .

 

 Dimensional Analysis and Volume

Exercise \(\PageIndex{2}\)

Solve these problems about lumber dimensions.

1. To describe to a European how houses are constructed in the US, the dimensions of “two-by-four” lumber must be converted into metric units. The thickness  ×  width  × length dimensions are 1.50 in.  ×  3.50 in.  ×  8.00 ft in the US. What are the dimensions in cm  ×  cm  ×  m?
2. This lumber can be used as vertical studs, which are typically placed 16.0 in. apart. What is that distance in centimeters?

Answer a

\(1 \;cm\;=\;2.54\;in\\1\;ft\;=\;.3048\;m\\1.50\textcolor{red}{ \cancel{in}}( \frac{2.54 \; cm}{1 \;\textcolor{red}{ \cancel{in}}})=3.81\;cm\\3.50\textcolor{red}{ \cancel{in}}( \frac{2.54 \; cm}{1 \;\textcolor{red}{ \cancel{in}}})=8.89\;cm\\8.00\textcolor{red}{ \cancel{ft}}( \frac{.3048 \; m}{1 \;\textcolor{red}{ \cancel{ft}}})=2.44\;m\\3.81\;cm\;\times\;8.89\;cm\;\times\;2.44\;m\)

The answer is 3.81 cm × 8.89 cm × 2.44 m (section 1B.4.2)

 

 

Answer b

\(1 \;cm\;=\;2.54\;in\\16.0\textcolor{red}{ \cancel{in}}( \frac{2.54 \; cm}{1 \;\textcolor{red}{ \cancel{in}}})=40.6\;cm\)

The answer is 40.6 cm (section 1B.4.2)

 

Exercise \(\PageIndex{3}\)

Calculate these volumes.

1. What is the volume of 11.3 g graphite in meters cubed, density = 2.25 g/cm3?
2. What is the volume of 39.657 g bromine in liters, density = 2.928 g/cm3?

Answer a

\(D=\frac{m}{V} \\ V=\frac{m}{D}\\V=\frac{11.3g}{2.25\frac{g}{cm^3}}\\V=5.02 cm^3\\V=\left (5.02 cm^3  \right )\left ( \frac{1m}{100cm^3} \right )^{3}=5.02\times 10^{-6}\;m\)

5.02 × 10^-6 m³ (section 1B.4.2)

 

Answer b

 

\(D=\frac{m}{V} \\ V=\frac{m}{D}\\V=\frac{39.657g}{2.928\frac{g}{cm^3}}\\V=13.54 cm^3\\V=\left (13.54 cm^3  \right )\left ( \frac{1L}{1000cm^3} \right )^{3}=5.02\times 10^{-2}\;L\)

1.354 ×10^-2 L (section 1B.4.2)

 

Temperature Conversions

Exercise \(\PageIndex{1}\)

Convert the following:

1. 37.0 ºC to _____ K
2. 331 K to _____ ºC

Answer a

310.2 K

\[(37.0+273.15)=310.2K\]

(see section 1B.1.5)

Answer b

57.9 ºC

\begin{equation}
(331-273.15)=57.9^{\circ} \mathrm{C}
\end{equation}

(see section 1B.1.5)

Exercise \(\PageIndex{2}\)

Convert the following:

1. 57.8 ºC to _____ ºF
2. -129 ºF to _____ ºC

Answer a

136 ºF

\begin{equation}
[(57.8+40) 1.8]-40=136^{\circ} \mathrm{F}
\end{equation}

(see section 1B.1.5)

Answer b

-89.4 ºC

\begin{equation}
\left(\frac{(-129+40)}{1.8}\right)-40=-89.4^{\circ} \mathrm{C}
\end{equation}

(see section 1B.1.5)

Exercise \(\PageIndex{3}\)

Convert the boiling temperature of gold, 2966 °C, into degrees Fahrenheit and Kelvin.

Answer

\(\left [\left (2966^{\circ}C+40  \right ) 1.8  \right ]-40=5371^{\circ}F\)

\(2966 ^{\circ} C +273.15 = 3239K\)

(see section 1B.1.5)

Exercise \(\PageIndex{4}\)

Convert the temperature of scalding water, 54 °C, into degrees Fahrenheit and Kelvin.

Answer

\(\left [\left (54^{\circ}C+40  \right ) 1.8  \right ]-40=130^{\circ}F\)

\(54 ^{\circ} C +273.15 = 330\)

130 °F, 330 K

(see section 1B.1.5)

Exercise \(\PageIndex{5}\)

Convert the temperature of the coldest area in a freezer, −10 °F, to degrees Celsius and Kelvin.

Answer

\( [(\frac{10 ^\circ F\;+\;40)}{1.6}]-40\;=\;-23 ^{\circ} C \)

\( [(\frac{10 ^\circ F\;+\;40)}{1.6}]-40\;+\;273.15\;=\;250\;K\)

−23 °C, 250 K

(see section 1B.1.5)

 

Algebra Review

 

Exercise \(\PageIndex{1}\)

Solve For \(A\)

\(A(X+Y)=Z\)

Answer

\(A(X+Y)=Z\\\frac{A(X+Y)}{X+Y}=\frac{Z}{X+Y}\\\frac{A\cancel{(X+Y)}}{\cancel{X+Y}}=\frac{Z}{X+Y}\\A=\frac{Z}{X+Y}\)

Exercise \(\PageIndex{1}\)

Solve For \(A\)

\((AX+Y)=Z\)

Answer

\((AX+Y)=Z\\(AX+Y)-Y=Z-Y\\(AX+\cancel{Y})-\cancel{Y}=Z-Y\\\frac{AX}{X}=\frac{Z-Y}{X}\\\frac{A\cancel{X}}{\cancel{X}}=\frac{Z-Y}{X}\\A=\frac{Z-Y}{X}\)

Exercise \(\PageIndex{1}\)

Solve For \(A\)

\((AX+AY)=Z\)

Answer

\((AX+AY)=Z\\A(X+Y)=Z\\\frac{A(X+Y)}{(X+Y)}=\frac{Z}{X+Y}\\\frac{A\cancel{(X+Y)}}{\cancel{(X+Y)}}=\frac{Z}{X+Y}\\A=\frac{Z}{X+Y}\)

Exercise \(\PageIndex{1}\)

Solve For \(A\)

\((AX+Y)Z=1\)

Answer

\((AX+Y)Z=1\\\frac{(AX+Y)Z}{Z}=\frac{1}{Z}\\\frac{(AX+Y)\cancel{Z}}{\cancel{Z}}=\frac{1}{Z}\\AX+Y-Y=\frac{1}{Z}-Y\\AX+\cancel{Y}-\cancel{Y}=\frac{1}{Z}-Y\\\frac{AX}{X}=\frac{\frac{1}{Z}-Y}{X}\\\frac{A\cancel{X}}{\cancel{X}}=\frac{\frac{1}{Z}-Y}{X}\\A=\frac{\frac{1}{Z}-Y}{X}\)

Exercise \(\PageIndex{1}\)

Solve For \(A\)

\(\frac{A}{X+Y}=Z\)

Answer

\(\frac{A}{X+Y}=Z\\\frac{A}{X+Y}(X+Y)=Z(X+Y)\\\frac{A}{\cancel{X+Y}}\cancel{(X+Y)}=Z(X+Y)\\A=Z(X+Y)\)

Exercise \(\PageIndex{1}\)

Solve For \(A\)

\(\frac{1}{AX+AY}=Z\)

Answer

\(\frac{1}{AX+AY}=Z\\AX+AY=\frac{1}{Z}\\A(X+Y)=\frac{1}{Z}\\\frac{A(X+Y)}{(X+Y)}=\frac{1}{Z(X+Y)}\\\frac{A\cancel{(X+Y)}}{\cancel{(X+Y)}}=\frac{1}{Z(X+Y)}\\A=\frac{1}{Z(X+Y)}\)

Exercise \(\PageIndex{1}\)

Solve For \(A\)

\(\frac{1}{X+AY}=Z\)

Answer

\(\frac{1}{X+AY}=Z\\X+AY=\frac{1}{Z}\\X+AY-X=\frac{1}{Z}-X\\\cancel{X}+AY\cancel{-X}=\frac{1}{Z}-X\\\frac{AY}{Y}=\frac{\frac{1}{Z}-X}{Y}\\\frac{A\cancel{Y}}{\cancel{Y}}=\frac{\frac{1}{Z}-X}{Y}\\A=\frac{\frac{1}{Z}-X}{Y}\)

Exercise \(\PageIndex{1}\)

Solve For \(A\)

\(\frac{A}{X+Y}=ZA+3\)

Answer

\(\frac{A}{X+Y}=ZA+3\\\frac{A}{X+Y}(X+Y)=(ZA+3)(X+Y)\\\frac{A}{\cancel{(X+Y)}}\cancel{(X+Y)}=(ZA+3)(X+Y)\\A=ZA(X+Y)+3(X+Y)\\A-ZA(X+Y)=ZA(X+Y)-ZA(X+Y)+3(X+Y)\\A-ZA(X+Y)=\cancel{ZA(X+Y)}-\cancel{ZA(X+Y)}+3(X+Y)\\    A(1-ZA(X+Y))=3(X+Y)\\\frac{A(1-ZA(X+Y))}{(1-ZA(X+Y))}=\frac{3(X+Y)}{(1-ZA(X+Y))}\\\frac{A\cancel{(1-ZA(X+Y))}}{\cancel{(1-ZA(X+Y))}}=\frac{3(X+Y)}{(1-ZA(X+Y))}\\A=\frac{3(X+Y)}{1-Z(X+Y)}\)

Exercise \(\PageIndex{1}\)

Solve For \(A\)

\(\frac{A}{X+Y}=ZA+D\)

Answer

\(\frac{A}{X+Y}=ZA+D\\\frac{A}{X+Y}(X+Y)=(ZA+D)(X+Y)\\\frac{A}{\cancel{(X+Y)}}\cancel{(X+Y)}=(ZA+D)(X+Y)\\A=ZA(X+Y)+D(X+Y)\\A-ZA(X+Y)=ZA(X+Y)-ZA(X+Y)+D(X+Y)\\A-ZA(X+Y)=\cancel{ZA(X+Y)}-\cancel{ZA(X+Y)}+D(X+Y)\\    A(1-ZA(X+Y))=D(X+Y)\\\frac{A(1-ZA(X+Y))}{(1-ZA(X+Y))}=\frac{D(X+Y)}{(1-ZA(X+Y))}\\\frac{A\cancel{(1-ZA(X+Y))}}{\cancel{(1-ZA(X+Y))}}=\frac{D(X+Y)}{(1-ZA(X+Y))}\\A=\frac{D(X+Y)}{1-Z(X+Y)}\)

Exercise \(\PageIndex{1}\)

Solve For \(A\)

\(\frac{A+P}{X+Y}=ZA+D\)

Answer

\(\frac{A+P}{X+Y}=ZA+D\\\frac{A+P}{X+Y}(X+Y)=(ZA+D)(X+Y)\\\frac{A+P}{\cancel{(X+Y)}}\cancel{(X+Y)}=(ZA+D)(X+Y)\\A+P=ZA(X+Y)+D(X+Y)\\A+P-P=ZA(X+Y)+D(X+Y)-P\\A+\cancel{P}-\cancel{P}=ZA(X+Y)+D(X+Y)-P\\A-ZA(X+Y)=ZA(X+Y)-ZA(X+Y)+D(X+Y)-P\\A-ZA(X+Y)=\cancel{ZA(X+Y)}-\cancel{ZA(X+Y)}+D(X+Y)-P\\    A(1-ZA(X+Y))=D(X+Y)-P\\\frac{A(1-ZA(X+Y))}{(1-ZA(X+Y))}=\frac{D(X+Y)-P}{(1-ZA(X+Y))}\\\frac{A\cancel{(1-ZA(X+Y))}}{\cancel{(1-ZA(X+Y))}}=\frac{D(X+Y)-P}{(1-ZA(X+Y))}\\A=\frac{D(X+Y)-P}{1-Z(X+Y)}\)

Text

1B.1: Units of Measurement

1B.2: Making Measurements: Experimental Error, Accuracy, Precision, Standard Deviation, and Significant Figures

1B.3: Mathematics in Chemistry

1B.4: Problem Solving by Dimensional Analysis

1B.5: Graphs and Graphing