Lesson 1.4 · Unit 1 · HS-PS1-7
SI units and dimensional analysis, the method that prevents most wrong answers
Chemistry is the first science most students meet in which a correct method and a wrong unit produce a wrong answer worth no marks. The defense is a procedure rather than care: write every quantity with its unit, multiply by conversion factors arranged so that unwanted units cancel on the page, and read the surviving unit at the end. If the surviving unit is not the one asked for, the setup is wrong and you know it before you touch the calculator.
- The SI base units you need in this course are the meter, the kilogram, the second, the kelvin and the mole. Everything else is derived from them, including the liter and the joule.
- Prefixes multiply by powers of ten: kilo \( 10^{3} \), centi \( 10^{-2} \), milli \( 10^{-3} \), micro \( 10^{-6} \), nano \( 10^{-9} \). Learn these five and most conversions become mechanical.
- A conversion factor is a fraction equal to one. Because \( 1000 \text{ mL} = 1 \text{ L} \), both \( \frac{1000 \text{ mL}}{1 \text{ L}} \) and \( \frac{1 \text{ L}}{1000 \text{ mL}} \) equal one, so multiplying by either changes the units without changing the quantity.
- Choose the orientation that cancels. Put the unit you want to remove in the denominator. This is the entire decision, and it removes the guessing about whether to multiply or divide.
- Chain as many factors as you need in one line. A conversion from gallons per day to liters per second is one expression, not three separate calculations, and keeping it as one line means one rounding at the end.
- Two volume equalities are worth memorizing: \( 1 \text{ mL} = 1 \text{ cm}^3 \) and \( 1 \text{ L} = 1000 \text{ cm}^3 \). Chemistry moves between these constantly.
Where students lose marks: leaving a volume in milliliters in a formula that requires liters. Molarity, the gas laws and every solution calculation in this course assume liters. Convert first, before the value goes anywhere near an equation.
The problem. A municipal water report gives the output of a treatment plant as 4.50 million gallons per day. Express this in liters per second. Use \( 1 \text{ gallon} = 3.785 \text{ L} \).
Step one: write the starting quantity with its full unit. \( 4.50 \times 10^{6} \; \frac{\text{gal}}{\text{day}} \). Both parts of the unit have to change, which is the only thing that makes this harder than a single conversion.
Step two: decide what must cancel. Gallons must go from the numerator, so the gallon conversion factor needs gallons in its denominator. Days must go from the denominator, so the time factors need days in their numerator.
Step three: build the chain.
\[ 4.50 \times 10^{6} \; \frac{\text{gal}}{\text{day}} \times \frac{3.785 \text{ L}}{1 \text{ gal}} \times \frac{1 \text{ day}}{24 \text{ h}} \times \frac{1 \text{ h}}{60 \text{ min}} \times \frac{1 \text{ min}}{60 \text{ s}} \]Step four: check the units before calculating. Gallons cancel against gallons, days against days, hours against hours, minutes against minutes. What survives is liters in the numerator and seconds in the denominator, which is what was asked for. The setup is correct, and this check costs five seconds.
Step five: do the numerator. \( 4.50 \times 10^{6} \times 3.785 = 1.70325 \times 10^{7} \) L per day.
Step six: do the denominator. \( 24 \times 60 \times 60 = 86\,400 \) seconds in a day. Converting the time in one step rather than three is fine once you trust the chain.
Step seven: divide and round. \( 1.70325 \times 10^{7} \div 86\,400 = 197.14 \). The starting value has three significant figures, so the answer is 197 L/s.
Step eight: sanity check the size. Roughly four and a half million gallons is about seventeen million liters, spread over about ninety thousand seconds, which should give something in the low hundreds. It does. An answer of 197 000 or 0.197 would have been caught here.
Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.
- Convert 2.5 kg to milligrams.
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\( 2.5 \text{ kg} \times \frac{1000 \text{ g}}{1 \text{ kg}} \times \frac{1000 \text{ mg}}{1 \text{ g}} \). \( 2.5 \times 10^{6} \) mg
- Convert 750 mL to liters.
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Divide by 1000. 0.750 L
- State the two volume equalities linking milliliters, cubic centimeters and liters.
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\( 1 \text{ mL} = 1 \text{ cm}^3 \) and \( 1 \text{ L} = 1000 \text{ cm}^3 \)
- Explain in one sentence why a conversion factor can be written either way up.
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Because the numerator and denominator are equal quantities, so the fraction equals one and multiplying by it cannot change the quantity
- Convert 85 km/h to meters per second.
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\( 85 \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = \frac{85\,000}{3600} \). 23.6 m/s
- A student converts 2.50 mL of mercury, density 13.53 g/cm3, to a mass in
kilograms. Carry out the conversion and explain where the two traps are.
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First trap: the density is per cubic centimeter and the volume is in milliliters, which looks like a mismatch but is not, because \( 1 \text{ mL} = 1 \text{ cm}^3 \) exactly. Students often insert a spurious conversion here. So \( 2.50 \text{ cm}^3 \times 13.53 \; \frac{\text{g}}{\text{cm}^3} = 33.825 \text{ g} \). Second trap: the answer is wanted in kilograms, and stopping at grams is the commoner error than getting the arithmetic wrong. Divide by 1000 to get 0.033825 kg, then round to the three significant figures the data support. 0.0338 kg
- Explain why checking that the units cancel is more useful than checking the
arithmetic.
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Because the two failures are not equally likely or equally damaging. Arithmetic slips are caught by redoing the calculation, and a calculator rarely mismultiplies. Setting a conversion factor the wrong way up produces an answer that is wrong by a factor of the conversion squared, often by a thousand or a million, and it looks entirely plausible on the page because the numbers are the right numbers. The unit check catches exactly that error, before any arithmetic is done, because an inverted factor leaves an impossible surviving unit such as gallons squared per liter. It also catches a missing step, since the wanted unit simply fails to appear. It catches inverted or missing factors, which produce large plausible errors that rechecking arithmetic will not find
- A laboratory procedure calls for 0.250 L of solution and a student measures 250 mL
in a graduated cylinder. A second student says this is wrong because the units
differ. Settle the disagreement.
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The measured quantity is correct, because 0.250 L and 250 mL are the same volume: multiplying 0.250 L by 1000 mL per liter gives 250 mL exactly. The second student has confused a difference in unit with a difference in quantity. There is a real point buried in the objection, though, and it concerns significant figures rather than size: 0.250 L is stated to three significant figures, so the measurement should be made in a vessel capable of supporting that precision. A 250 mL graduated cylinder typically reads to about one milliliter, which is adequate, while a 250 mL beaker's graduations are not. The volumes are identical; the defensible concern is precision, not magnitude
- A carbon dioxide concentration is reported as 421 parts per million by volume.
Express this as a percentage by volume, and comment on why the smaller unit is used.
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Parts per million means parts per \( 10^{6} \), and percent means parts per \( 10^{2} \), so divide by \( 10^{4} \): \( 421 \div 10^{4} = 0.0421 \) percent. The reason the atmospheric record is published in parts per million rather than percent is that the interesting changes are invisible at the coarser unit. A rise from 280 to 421 parts per million over two centuries appears as a change from 0.028 to 0.042 percent, which reads as almost nothing and requires three decimal places to express at all, whereas the same change in parts per million is a rise of 141 in a quantity near 300. The unit should be chosen so that ordinary variation lands in the ones or tens place. 0.0421 percent; parts per million is used so the variation is legible without leading zeros
- Explain why every equation in this course that involves concentration or gas volume
assumes liters, and what that implies for your working habits.
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The constants are defined that way. Molarity is defined as moles per liter, and the gas constant \( R \) has the value 0.0821 with units of liter atmospheres per mole kelvin, so a volume substituted in milliliters is wrong by a factor of a thousand and nothing in the equation reveals it. The result looks like a normal number and carries no warning. The working habit that prevents this is to convert units at the point of reading the question rather than at the point of substitution: write the volume down in liters, the temperature in kelvin and the mass in grams as you extract each one, so the quantities entering any formula are already in the units it expects. The constants are defined per liter, so convert at the moment you read the data, not at substitution