Lesson 1.6 · Unit 1 · HS-PS1-7
Scientific notation, orders of magnitude, and estimating before you calculate
Chemistry routinely handles numbers separated by twenty-five orders of magnitude, from the mass of an atom to the number of particles in a spoonful. Ordinary decimal notation fails at both ends, and more importantly it hides errors: a calculator answer that is wrong by a factor of a thousand looks exactly as respectable as a right one. Estimating the order of magnitude first is the cheapest error check in the subject.
- Scientific notation writes a number as \( M \times 10^{n} \), where \( M \) is at least 1 and less than 10 and \( n \) is an integer. Only the digits in \( M \) are significant.
- A positive exponent means a number larger than one; a negative exponent means smaller than one. The exponent counts places moved, not zeros.
- To multiply, multiply the coefficients and add the exponents. To divide, divide the coefficients and subtract the exponents.
- To add or subtract, the exponents must match first. Rewrite one number so both share an exponent, then add the coefficients.
- Fix the coefficient afterward. If the result is \( 43 \times 10^{4} \), rewrite it as \( 4.3 \times 10^{5} \). An answer left with a coefficient outside the range is not in scientific notation.
- An order of magnitude is a factor of ten. Two quantities differing by three orders of magnitude differ by a factor of a thousand, and saying so is often more useful than either exact value.
- Estimate first. Round every value to one digit, do the arithmetic mentally, and you have a range the real answer must fall in.
Where students lose marks: adding the exponents when the operation was addition. The exponent rules apply to multiplication and division only; addition requires matching exponents first, and the exponent then stays the same.
The problem. A typical atom has a diameter of about \( 1 \times 10^{-10} \) m. Roughly how many atoms would lie in a line across the thickness of a credit card, about 0.76 mm? Then say what that number tells you.
Step one: put both quantities in the same unit and in scientific notation. The card is 0.76 mm, which is \( 0.76 \times 10^{-3} \) m. The coefficient must be at least one, so rewrite this as \( 7.6 \times 10^{-4} \) m.
Step two: state the calculation. The number of atoms across is the total length divided by the length of one atom:
\[ N = \frac{7.6 \times 10^{-4} \text{ m}}{1 \times 10^{-10} \text{ m}} \]Step three: estimate before calculating. Roughly \( 10^{-4} \) divided by \( 10^{-10} \) is \( 10^{6} \), so the answer should be in the millions. If the calculator returns something in the thousands, a sign on an exponent went wrong.
Step four: divide the coefficients. \( 7.6 \div 1 = 7.6 \).
Step five: subtract the exponents. \( -4 - (-10) = -4 + 10 = 6 \). Subtracting a negative exponent is where sign errors appear, so write the double negative out rather than doing it in your head.
Step six: assemble and check the units. \( N = 7.6 \times 10^{6} \). Meters cancel against meters, leaving a pure count, which is correct for "how many atoms". It agrees with the estimate.
Step seven: say what the number means. About seven and a half million atoms span a card that you would describe as thin. This is why chemistry cannot work by counting particles individually and needs the mole, which unit 4 introduces: the gap between the scale we measure at and the scale reactions happen at is six or more orders of magnitude in length and twenty-three in number.
Step eight: note the precision honestly. The atomic diameter was given as "about" one ten-billionth of a meter and atoms vary by a factor of about three across the periodic table, so this answer is good to one significant figure at best. Writing 7 600 000 would overstate it. The right claim is "of the order of ten million".
Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.
- Write 0.000 062 5 in scientific notation.
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Move the decimal point five places right. \( 6.25 \times 10^{-5} \)
- Write \( 4.07 \times 10^{4} \) in ordinary decimal notation.
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40 700
- Calculate \( (3.0 \times 10^{8}) \times (2.0 \times 10^{-5}) \).
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Coefficients \( 3.0 \times 2.0 = 6.0 \); exponents \( 8 + (-5) = 3 \). \( 6.0 \times 10^{3} \)
- Calculate \( (6.0 \times 10^{5}) \div (2.0 \times 10^{8}) \).
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Coefficients \( 6.0 \div 2.0 = 3.0 \); exponents \( 5 - 8 = -3 \). \( 3.0 \times 10^{-3} \)
- How many orders of magnitude separate \( 2 \times 10^{3} \) and
\( 2 \times 10^{9} \)?
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The exponents differ by six. Six, a factor of a million
- A student calculates \( (4.0 \times 10^{3}) + (2.0 \times 10^{2}) \) and answers
\( 8.0 \times 10^{5} \). Find every error and give the correct result.
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They applied the multiplication rules to an addition, multiplying the coefficients and adding the exponents. Neither operation is permitted here. For addition the exponents must match first, so rewrite \( 2.0 \times 10^{2} \) as \( 0.20 \times 10^{3} \). Now the coefficients can be added: \( 4.0 + 0.20 = 4.2 \), and the shared exponent is unchanged, giving \( 4.2 \times 10^{3} \). A check in ordinary notation confirms it: \( 4000 + 200 = 4200 \). Their answer of 800 000 is nearly two hundred times too large, which an order-of-magnitude estimate would have caught immediately. \( 4.2 \times 10^{3} \)
- Explain why scientific notation makes significant figures unambiguous, using 500 as
the example.
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Written as 500, the two zeros might be measured digits or might be placeholders fixing the magnitude, and nothing in the notation says which, so the value could be claiming one, two or three significant figures. Scientific notation separates the two jobs completely: the power of ten carries all the magnitude information and the coefficient carries all the precision information, so every digit written in the coefficient is by definition significant. \( 5 \times 10^{2} \) claims one significant figure, \( 5.0 \times 10^{2} \) claims two and \( 5.00 \times 10^{2} \) claims three, and a reader knows exactly which measurement was made. Magnitude sits in the exponent and precision in the coefficient, so every written digit is significant
- A calculation of the number of molecules in a drop of water returns
\( 1.7 \times 10^{21} \). Another student's answer is \( 1.7 \times 10^{12} \).
Explain how you could tell which is plausible without redoing the calculation.
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By comparing each against a fixed landmark. A mole is \( 6.02 \times 10^{23} \) particles and corresponds to about 18 g of water, so a drop of roughly 0.05 g is a few thousandths of a mole, which must therefore contain something in the region of \( 10^{21} \) molecules. The first answer sits exactly there. The second is nine orders of magnitude smaller, which would mean a drop of water contained about a millionth of a millionth of a mole, or a mass far below what any balance would register. Holding one or two anchor values in mind, such as Avogadro's number and the molar mass of water, lets you reject an answer by scale alone. The first; comparing with Avogadro's number shows the second is nine orders of magnitude too small
- Explain why the answer to the worked example should be stated as "of the order of
ten million" rather than as 7 600 000.
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Because the input did not support seven digits. The atomic diameter was given as about \( 1 \times 10^{-10} \) m, which is a single significant figure and is in any case a rough average, since atomic diameters range by roughly a factor of three from helium to cesium. An answer can never be more precise than the worst input, so at most one significant figure is defensible here, and even that digit depends on which element the atoms are. Writing 7 600 000 implies the count was determined to the nearest hundred, which no part of the calculation could justify. Stating an order of magnitude reports exactly what the estimate supports. The input had one significant figure and atomic sizes vary, so only the order of magnitude is supported
- An atom is about \( 10^{-10} \) m across and a nucleus about \( 10^{-15} \) m.
Express the relationship as an order of magnitude and explain why Rutherford found
his result surprising.
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The diameters differ by five orders of magnitude, so the atom is about a hundred thousand times wider than its nucleus. Because volume scales with the cube of length, the nucleus occupies roughly \( 10^{15} \) times less volume than the atom, meaning the atom is overwhelmingly empty space. The surprise in Rutherford's gold foil experiment follows directly: on the prevailing model, with positive charge spread evenly through the atom, alpha particles should have passed through a thin foil with only slight deflections, and almost all of them did. The occasional particle returning nearly the way it came required all the positive charge and nearly all the mass to be concentrated in that tiny fraction of the volume, which lesson 2.2 takes up. Five orders of magnitude in diameter and about fifteen in volume, so the atom is nearly all empty space