Quantify matter with SI units, significant figures, average atomic mass and the mole

Learning goal: Report laboratory measurements in SI units with the correct number of significant figures, distinguish accuracy from precision, calculate average atomic mass from isotope abundances, use the mole and molar mass to convert between mass, moles, number of particles and volume by dimensional analysis, and find percent composition and an empirical formula from data.

Three domains of size. Chemistry works across three domains of magnitude: the macroscopic domain of grams and milliliters that you can measure directly, the cosmic domain of stars and galaxies, and the submicroscopic domain of atoms and molecules. Only the macroscopic domain is reached with ordinary tools. Measurements in the other two need complex instruments or indirect methods, so much of quantitative chemistry is about linking a macroscopic measurement, a mass on a balance, to a submicroscopic count, the number of particles. That link is the mole.

SI units and scientific notation. Scientific communication uses the International System of Units: the kilogram for mass (the gram in practice), the meter and its derived liter for volume (1 L = 1 dm³ = 1000 cm³ = 1000 mL), the kelvin for temperature, the second for time, and the mole for amount of substance. Very large and very small numbers are written in scientific notation: a coefficient between 1 and 10 times a power of ten. Avogadro's number is 6.022 × 10²³, and a single carbon atom has a mass of about 1.99 × 10⁻²³ g. The ratio between an atom about 10⁻¹⁰ m across and a person about 1.7 m tall is roughly 10¹⁰, the same ratio as between a person and a distance of 17 000 000 km, more than forty times the distance to the Moon; that comparison is the point of scientific notation.

Accuracy and precision are different. Every measurement carries error. Accuracy is how close a measurement is to the true or accepted value; precision is how closely repeated measurements agree with one another. A balance that reads 24.9, 25.0 and 25.1 g for a 30.0 g standard is precise but not accurate, probably because it was not zeroed; a balance that reads 31, 35 and 33 g is neither. (Readings of 28, 32 and 30 g would be imprecise but accurate on average, because their mean is 30.) Instruments differ in precision: a 100 mL beaker's markings are guides good to perhaps 5 mL, a 100 mL graduated cylinder can be read to 0.5 mL, and a burette to 0.05 mL. The correct tool is the one whose precision matches the need; a beaker is fine for rinse water and wrong for a titration.

Significant figures. The significant figures in a measurement are the digits known with certainty plus one estimated digit, and they tell the reader how precise the instrument was. Rules for counting: all nonzero digits are significant; zeros between nonzero digits are significant (2.05 has three); leading zeros are not (0.0025 has two); trailing zeros are significant only if there is a decimal point (25.0 has three; 250 with no decimal point is ambiguous: many textbooks and tests count such zeros as not significant, giving two, but the writer may have measured to three, so write 2.5 × 10², 2.50 × 10² or 250. with a decimal point to say which). Rules for operations: in multiplication and division the answer keeps as many significant figures as the measurement with the fewest; in addition and subtraction the answer keeps as many decimal places as the measurement with the fewest. Exact numbers, such as the 2 in two hydrogens per water molecule or 1000 mL in a liter, never limit an answer. The reason for the rules is honesty: a calculator shows 0.001530864 but your data do not know that many digits.

Dimensional analysis. Every conversion in this course is done by multiplying by a ratio equal to one, written so that the unit you have cancels and the unit you want remains. To convert 2.50 L to milliliters, multiply by 1000 mL / 1 L: 2.50 L × (1000 mL / 1 L) = 2500 mL, written 2.50 × 10³ mL to keep three significant figures. Writing the units and canceling them is not decoration; it is the check that the ratio was set up the right way up.

Average atomic mass. An element in nature is a mixture of isotopes, and the atomic mass on the periodic table is the average of their masses weighted by their natural abundance. Multiply each isotope's mass by its fractional abundance and add. Copper, with isotope masses in atomic mass units (u): 0.6917 × 62.93 + 0.3083 × 64.93 = 43.53 + 20.02 = 63.55 u. (The same number with the unit grams per mole is copper's molar mass, introduced next.) Working backwards, if an element's atomic mass sits close to one isotope's mass number, that isotope is the more abundant one.

The mole. Because atoms are so small, chemists count them in a large fixed group. One mole is 6.022 × 10²³ particles; historically it was defined as the number of atoms in exactly 12 g of carbon-12, and since 2019 the count itself has been fixed by definition, which changes nothing at the precision used here. The value is Avogadro's number. The mass of one mole of a substance is its molar mass in grams per mole, and it is numerically equal to the atomic mass for an element or the sum of the atomic masses in the formula for a compound. Using atomic masses to two decimals: water H₂O is 2(1.01) + 16.00 = 18.02 g/mol; carbon dioxide is 12.01 + 2(16.00) = 44.01 g/mol; calcium carbonate CaCO₃ is 40.08 + 12.01 + 3(16.00) = 100.09 g/mol; glucose C₆H₁₂O₆ is 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol. The mole translates between the domains: a balance reads grams, molar mass turns grams into moles, and Avogadro's number turns moles into particles.

A worked model: three conversions from one map. Moles are the hub. Mass to moles: divide by molar mass. Moles to particles: multiply by 6.022 × 10²³. Volume to mass for a liquid or solid: multiply by density. Suppose you have 10.0 g of calcium carbonate. Moles: 10.0 g × (1 mol / 100.09 g) = 0.0999 mol. Formula units: 0.0999 mol × (6.022 × 10²³ / 1 mol) = 6.02 × 10²² formula units. Now a liquid: 25.0 mL of ethanol, density 0.789 g/mL, molar mass C₂H₆O = 2(12.01) + 6(1.01) + 16.00 = 46.08 g/mol. Mass: 25.0 mL × 0.789 g/mL = 19.7 g. Moles: 19.7 g / 46.08 g/mol = 0.428 mol. Each answer is limited to three significant figures by the data. Volumes of gases convert through the molar volume, which belongs to the gas laws lesson.

Percent composition and empirical formula. The percent by mass of an element in a compound is that element's total mass in one mole divided by the molar mass, times 100. In water hydrogen is 2.02 / 18.02 = 11.2 percent and oxygen 88.8 percent. In the hydrate copper(II) sulfate pentahydrate, CuSO₄·5H₂O, molar mass 159.62 + 5(18.02) = 249.72 g/mol, the water is 90.10 / 249.72 = 36.08 percent, which is what a heating experiment should find as mass loss. An empirical formula is the simplest whole-number ratio of atoms and is found by turning percentages into moles: a compound that is 40.0 percent carbon, 6.7 percent hydrogen and 53.3 percent oxygen has, per 100 g, 40.0 / 12.01 = 3.33 mol C, 6.7 / 1.01 = 6.6 mol H and 53.3 / 16.00 = 3.33 mol O; dividing by the smallest gives 1 : 2 : 1, so the empirical formula is CH₂O. Glucose, C₆H₁₂O₆, has this empirical formula with every subscript multiplied by 6; the empirical formula alone cannot tell you which multiple is right.

Moles or mass? Mass is what you measure and what a recipe or a shipping label needs; moles are what a reaction counts. Identify the situations where each is most appropriate to use: when comparing amounts that will react, moles are the right unit, because equations are written in particles; when weighing out a sample, buying a chemical or reporting a yield, mass is the natural unit, because a balance and a price list work in grams. Density, mass per unit volume, is a property of a substance and is the same for a drop as for a bucket; a graph of mass against volume for several samples of one uniform liquid, all at the same temperature, is a straight line through the origin whose slope is the density (a different liquid, a mixture of a different strength or a different temperature gives a different slope).

Investigation: measure what a ruler cannot. Materials: a rectangle of aluminum foil, a ruler, a balance reading to 0.01 g, a graduated cylinder, water, an irregular small object such as a bolt. Part A: measure the length and width of the foil to the nearest millimeter, find its mass, and use the density of aluminum, 2.70 g/cm³, to find the volume and then the thickness, which no ruler can read directly. Part B: read the cylinder's water level, submerge the bolt, read again, and report the volume in cm³ to the precision of the cylinder; find the bolt's mass and its density with the correct significant figures. Record every reading with its unit and its estimated digit. Part C: measure 50 mL of water three times each with a beaker, a graduated cylinder and, if available, a burette or pipette, weighing each portion; compare the accuracy of each measuring device by how close the mass comes to 50.0 g and its precision by how much the three readings scatter. A result that supports the claim: the foil thickness comes out around 10 to 25 micrometers, far below the ruler's resolution, and the bolt's density is close to the value for steel; a thickness or density with the wrong number of significant figures is not a correct result. What the model cannot show: the foil calculation assumes the sheet is uniform and pure aluminum, and displacement cannot measure a porous or floating object.

Worked example 1 — foil thickness with significant figures. A sheet of foil is 10.0 cm by 15.0 cm and has a mass of 0.62 g. Area: 10.0 × 15.0 = 150 cm² (three significant figures). Volume from density: 0.62 g / 2.70 g/cm³ = 0.2296 cm³, but the mass has only two significant figures, so the volume is 0.23 cm³. Thickness = volume / area = 0.2296 / 150 = 0.00153 cm before rounding, which is 1.5 × 10⁻³ cm or 15 micrometers to two significant figures. Carry extra digits through the calculation and round only at the end.

Worked example 2 — mass to particles. How many molecules are in 4.50 g of water? Molar mass 18.02 g/mol. Moles: 4.50 g × (1 mol / 18.02 g) = 0.2497 mol, which is 0.250 mol to three significant figures. Molecules: carrying the unrounded 0.2497 mol, 0.2497 × (6.022 × 10²³ molecules / 1 mol) = 1.50 × 10²³ molecules. Had you rounded to 0.250 first, you would get 1.51 × 10²³, which is why the rule is to round only at the end. Reversed: 0.250 mol of water has a mass of 0.250 × 18.02 = 4.51 g; the small difference from 4.50 is rounding in the 0.250, a reminder that rounding intermediate values introduces error.

Worked example 3 — significant figures in operations. Three masses are added: 12.11 g + 0.3 g + 1.234 g = 13.644 g on the calculator. The least precise measurement, 0.3 g, has one decimal place, so the sum is 13.6 g. Then a density: 13.6 g / 5.2 mL = 2.615 g/mL on the calculator; the volume has two significant figures, so the density is 2.6 g/mL. Addition follows decimal places; division follows significant figures.

Worked example 4 — percent water in a hydrate from data. A hydrate sample of mass 5.00 g is heated gently (with a supervising adult) until its mass stops changing, at 3.20 g. Water lost: 5.00 − 3.20 = 1.80 g. Percent water: 1.80 / 5.00 × 100 = 36.0 percent. The calculated value for CuSO₄·5H₂O is 36.08 percent, so the data are consistent with that formula. If the sample were heated too strongly and the copper sulfate itself decomposed, the mass loss would overshoot and the percent would be too high; that is a systematic error, not a random one.

Representation task: the mole map. Draw a diagram with moles in the center and four boxes around it: mass (grams), number of particles, volume of a liquid or solid, and volume of a solution (for later). Write on each arrow the conversion factor and its direction: molar mass, Avogadro's number, density. Under the map, work these three and show every unit canceling: 2.00 mol of CO₂ to grams; 3.01 × 10²³ atoms of copper to moles and grams; 50.0 mL of water (density 1.00 g/mL) to moles.

Independent performance: measurement report with error analysis. Carry out the foil and displacement investigation and write it up. Success criteria: every raw reading recorded with unit and the correct number of significant figures for the instrument; the foil thickness in centimeters and micrometers with two significant figures and the dimensional analysis shown; the bolt's density with correct significant figures; a paragraph distinguishing accuracy from precision using your own repeated readings; a calculation of the average atomic mass of an element from isotope data your teacher supplies; and an empirical formula from a percent composition your teacher supplies, with the mole ratio shown.

Extension: percent water in popcorn. Weigh a counted number of kernels, pop them with an adult in a microwave popcorn popper made for that purpose (following its instructions, no oil, and stopping when pops are more than two seconds apart, because unattended popcorn can scorch or ignite), let them cool, and weigh again. The mass lost is mostly water that escaped as steam, so the loss divided by the starting mass, times 100, estimates the percent by mass of water lost during popping; it is an estimate, not the kernels' total water content, because some water may stay behind and bits of hull may be lost. Compare three brands, keeping the popping method and time the same, and correlate the percent of water lost with the amount of popcorn produced, counted as popped kernels out of the number started. Present the findings in a table, a bar chart and a short written conclusion.