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Mathematical Skills in Chemistry

Units, standard form and proportion in chemical calculation: standard form

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Overview

Thirteen recall questions on units, standard form and proportion in chemical calculation: standard form.

What you need to know

Answer each one before you reveal it.

Your recallNot started
Numbers

Write 0.0050 mol dm⁻³ in standard form, and explain how many significant figures the value carries.

Key term

Convert 1.25 × 10⁻² mol dm⁻³ to ordinary form without altering the precision implied.

Numbers

A student converts 2.400 × 10³ g mol⁻¹ to ordinary form and writes 2400 g mol⁻¹. Why is this unsatisfactory?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Write 0.0050 mol dm⁻³ in standard form, and explain how many significant figures the value carries.

    Model answer

    0.0050 mol dm⁻³ = 5.0 × 10⁻³ mol dm⁻³. The leading zeros only fix the decimal point and are not significant, but the final zero is, so the value has two significant figures. Standard form makes this unambiguous.

  2. 2

    Convert 1.25 × 10⁻² mol dm⁻³ to ordinary form without altering the precision implied.

    Model answer

    1.25 × 10⁻² mol dm⁻³ = 0.0125 mol dm⁻³. Three significant figures are retained; writing 0.013 would discard information the original value carried.

  3. 3

    A student converts 2.400 × 10³ g mol⁻¹ to ordinary form and writes 2400 g mol⁻¹. Why is this unsatisfactory?

    Model answer

    The standard form value shows four significant figures, but 2400 written plainly is ambiguous because the trailing zeros could be placeholders. Standard form should be kept, or the precision stated explicitly, so that the reader knows the value is certain to the nearest gram per mole.

  4. 4

    Express the ionic product of water at 298 K in ordinary form and comment on why standard form is preferred.

    Model answer

    1.00 × 10⁻¹⁴ mol² dm⁻⁶ is 0.0000000000000100 mol² dm⁻⁶ in ordinary form. Standard form is preferred because it is compact, the number of significant figures is unambiguous, and the exponent alone communicates the order of magnitude.

  5. 5

    Calculate the number of molecules in 4.40 g of carbon dioxide. (Avogadro constant = 6.02 × 10²³ mol⁻¹, M(CO₂) = 44.0 g mol⁻¹)

    Model answer

    n = m ÷ M = 4.40 ÷ 44.0 = 0.100 mol. Number of molecules = 0.100 × 6.02 × 10²³ = 6.02 × 10²² molecules.

  6. 6

    Calculate the mass of a single molecule of water in grams. (M(H₂O) = 18.0 g mol⁻¹, L = 6.02 × 10²³ mol⁻¹)

    Model answer

    Mass of one molecule = 18.0 ÷ (6.02 × 10²³) = 2.99 × 10⁻²³ g. Dividing by a number in standard form subtracts the exponent, so the answer is very small, as expected for a single molecule.

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