Measurement, units, significant figures and the mole

Chemistry is a numbers course. Most of the questions on the midterm are of the type “this many grams are given, how much of this forms”, and all of them rely on the three tools of this week: unit conversion, significant figures and the mole. Think of them like a programmer’s view of a type system: every number has a unit, and if the units do not match, the calculation is wrong.

Golden rules:

1. Never drop the unit. Write every number with its unit and cancel the units. If the unit left at the end is not the one asked for, the calculation is wrong.
2. The bridge between grams and atoms is the mole. Grams → moles → particles. Without the mole there is no stoichiometry.
3. Significant figures are the honesty of a result. Do not write all 12 digits the calculator gives.
4. Read the formula: in $\mathrm{Ca(NO_3)_2}$ there are 2 N and 6 O. A subscript outside the parentheses multiplies everything inside.

SI units and prefixes

The International System of Units (SI units) is based on seven base units. The ones you use most in chemistry:

Volume is a derived unit: $\text{m}^3$. But in the laboratory the liter is used: $1\ \text{L} = 1\ \text{dm}^3 = 1000\ \text{cm}^3$ and $1\ \text{mL} = 1\ \text{cm}^3$.

Converting between temperature scales involves addition, not multiplication:

$T(\text{K}) = t(^\circ\text{C}) + 273.15 \qquad t(^\circ\text{C}) = \big(t(^\circ\text{F}) - 32\big)\times\tfrac{5}{9}$

In gas laws and thermodynamics always use kelvin. This rule will earn you a lot of points in week 4.

Uncertainty: precision and accuracy

Every measurement has an uncertainty. Do not confuse the two terms:

• Precision: how close repeated measurements are to each other.
• Accuracy: how close the measurements are to the true value.

Let the true value be 10.00 g. A balance that reads 9.51, 9.50, 9.52 g is precise but not accurate (there is a systematic error). A balance that reads 9.8, 10.3, 10.0 is correct on average but scattered.

Significant figures

In a measurement, the digits known for certain plus the last estimated digit are significant. The counting rules:

Resolve the ambiguity with scientific notation: $1.50\times10^3$ has three significant figures.

Exact numbers have infinitely many significant figures: counted objects (3 molecules), definitions (1 in = 2.54 cm, 1 km = 1000 m). They do not limit the result.

Rules in calculations

• Multiplication and division: the result has as many significant figures as the number with the fewest significant figures.
• Addition and subtraction: the result has as many decimal places as the number with the fewest decimal places.

4.52 × 0.031  = 0.14012   → 0.14       (0.031 has two significant figures)
15.02 + 9.1   = 24.12     → 24.1       (9.1 has one decimal place)
13.0 - 10.0   = 3.0       → 2 significant figures! (subtraction eats digits)

PITFALL · rounding in intermediate steps. In multi-step calculations keep the intermediate results in full and round only at the very end. If you round at every step, the last digit drifts.

Dimensional analysis

The only reliable way to convert units is to multiply by a conversion factor. A conversion factor is a fraction whose value is 1, for example $\frac{2.54\ \text{cm}}{1\ \text{in}}$. You write it so that the unit you want to discard is in the denominator and the unit you want is in the numerator.

Example 1. How many cm is 1.75 ft?

$1.75\ \text{ft} \times \frac{12\ \text{in}}{1\ \text{ft}} \times \frac{2.54\ \text{cm}}{1\ \text{in}} = 53.3\ \text{cm}$

12 and 2.54 are exact because they are definitions; the three significant figures of 1.75 decide the result.

Example 2. How many m/s is 55.0 mi/h? (1 mi = 1609.344 m)

$55.0\ \frac{\text{mi}}{\text{h}} \times \frac{1609.344\ \text{m}}{1\ \text{mi}} \times \frac{1\ \text{h}}{3600\ \text{s}} = 24.6\ \text{m/s}$

Example 3 (square and cubic units). How many kg/m³ is 3.20 g/cm³?

$3.20\ \frac{\text{g}}{\text{cm}^3} \times \frac{1\ \text{kg}}{1000\ \text{g}} \times \left(\frac{100\ \text{cm}}{1\ \text{m}}\right)^3 = 3.20\times10^{3}\ \frac{\text{kg}}{\text{m}^3}$

PITFALL · forgetting the cube. $1\ \text{m}^3 = 100^3\ \text{cm}^3 = 10^6\ \text{cm}^3$, not 100. Put the whole factor in parentheses and apply the exponent.

Density

$d = \frac{m}{V}$

Density is a conversion factor between mass and volume. The mass of 250.0 mL of ethanol ($d = 0.789$ g/mL) is:

$250.0\ \text{mL} \times \frac{0.789\ \text{g}}{1\ \text{mL}} = 197\ \text{g}$

Example 4 (water displacement). An 18.7 g piece of metal is dropped into a graduated cylinder, and the water level rises from 10.0 mL to 13.0 mL. The volume is $13.0 - 10.0 = 3.0$ mL (two significant figures). The density is $18.7 / 3.0 = 6.2$ g/mL. The calculator says 6.233, but the subtraction ate one significant figure.

The mole concept

Atoms are too small to be weighed one by one. That is why chemists use a “package” unit: the mole. One mole is $6.022\times10^{23}$ particles. This number is called Avogadro’s number ($N_A$).

The mole is useful because the molar mass ($M$, g/mol) of an element is the same number as the atomic mass in the periodic table. The atomic mass of carbon is 12.011 u, so 1 mol of C is 12.011 g.

The molar mass of a compound is the sum of the masses of the atoms in the formula:

$M(\mathrm{H_2SO_4}) = 2(1.008) + 32.06 + 4(15.999) = 98.07\ \text{g/mol}$

Three basic conversions:

$n = \frac{m}{M} \qquad N = n \times N_A \qquad m = n \times M$

   grams  --(÷ M)-->  moles  --(× N_A)-->  particles (molecules/atoms)
   grams  <--(× M)--  moles  <--(÷ N_A)--  particles
   subscript in the formula: mol of compound → mol of atoms

Example 5. How many moles, how many molecules and how many O atoms are in 50.0 g of $\mathrm{H_2SO_4}$?

$n = 50.0\ \text{g} \times \frac{1\ \text{mol}}{98.07\ \text{g}} = 0.510\ \text{mol}$

$N_{\text{molecules}} = 0.5098\ \text{mol} \times 6.022\times10^{23}\ \text{mol}^{-1} = 3.07\times10^{23}$

$N_{\text{O}} = 3.070\times10^{23} \times 4 = 1.23\times10^{24}$

Compounds and their formulas

• Molecular formula: how many atoms there really are in the molecule. Glucose is $\mathrm{C_6H_{12}O_6}$.
• Empirical formula: the simplest whole-number ratio of the atoms. For glucose it is $\mathrm{CH_2O}$.
• Ionic compounds have no molecules; a formula unit is written: $\mathrm{NaCl}$, $\mathrm{CaCl_2}$.

Percent composition

$\%\,\text{element} = \frac{(\text{number of atoms in the formula}) \times (\text{atomic mass})}{M_{\text{compound}}} \times 100$

For $\mathrm{Fe_2O_3}$, $M = 159.69$ g/mol: $\%\mathrm{Fe} = \frac{2(55.845)}{159.69}\times100 = 69.94\%$.

From percentages to the empirical formula

The recipe: assume 100 g, so the percentages become grams. Convert grams to moles, divide by the smallest number of moles, and round to integers.

Example 6. A compound contains 26.58% K, 35.35% Cr and 38.07% O by mass.

K : 26.58 g / 39.098 g/mol = 0.6798 mol   ÷ 0.6798 = 1.000
Cr: 35.35 g / 51.996 g/mol = 0.6799 mol   ÷ 0.6798 = 1.000
O : 38.07 g / 15.999 g/mol = 2.380 mol    ÷ 0.6798 = 3.501
ratio 1 : 1 : 3.5  → multiply all by 2 → 2 : 2 : 7

The empirical formula is $\mathrm{K_2Cr_2O_7}$ (potassium dichromate).

TIP · halves and thirds. If the ratio is 1.5, multiply by 2; if it is 1.33 or 1.67, multiply by 3; if it is 1.25, multiply by 4. Do not round 3.501 to “3”; deviations larger than 0.1 mean you need to multiply.

From the empirical formula to the molecular formula

If the molar mass is given: find the whole number $k = \dfrac{M_{\text{molecule}}}{M_{\text{empirical}}}$ and multiply every subscript in the empirical formula by $k$. For a compound with empirical formula $\mathrm{CH}$ (13.02 g/mol) and molar mass 78.1 g/mol, $k = 6$ and the formula is $\mathrm{C_6H_6}$.

Combustion analysis

A substance containing C, H (and O) is burned in excess oxygen. All the C goes to $\mathrm{CO_2}$; all the H goes to $\mathrm{H_2O}$. You work backward from the masses:

• $\text{mol C} = \text{mol CO}_2$
• $\text{mol H} = 2 \times \text{mol H}_2\text{O}$
• If O is present: $m_{\mathrm{O}} = m_{\text{sample}} - m_{\mathrm{C}} - m_{\mathrm{H}}$

Do not compute O from the products: since $\mathrm{O_2}$ is added from outside in the combustion, the O in the products does not come only from the sample.

Hydrates

Crystals like $\mathrm{CuSO_4\cdot 5H_2O}$ hold water in their structure. When heated the water escapes; the lost mass is water. The ratio mol water / mol anhydrous salt gives x in the formula.

Common mistakes