CHM 101/FSC 112: Errors, Accuracy, Precision

• CHM 101/FSC 112 covers topics including errors, accuracy and precision, and stoichiometry.
• The intended learning outcomes include being able to measure error, explain accuracy and precision, and apply the Mole concept to calculations.

Errors

• All measurements that involve a numerical answer have an error.
• The error can come from the instrument or personal error, such as an analyst being unable to read the instrument correctly.
• The error of a measurement defines the difference between measured and true values.
• The error is also called the absolute error.
• Absolute error = measured value - true value
• Errors can be expressed as relative error.
• Relative error = absolute error / true value
• Percentage relative error = (|absolute error| / true value) x 100
• For example, if a goldsmith measures 22.5 g of gold, but the actual weight is 25.2 g:
• The absolute error is 22.5 - 25.2 = -2.7.
• The relative error is -2.7 / 25.2 = -0.11.
• The percentage relative error (-2.7 / 25.2) x 100 = -11%.

Accuracy and Precision

• Accuracy and precision are important when making measurements.
• Accuracy is defined as the degree of agreement between the measured value and the true value.
• Precision is defined as the extent to which results agree with one another, or the repeatability of a result.
• For example, if the true value is 2.000g:
  • Student A measures results of 1.964 g, 1.978 g, and 1.960 g, with an average of 1.967 g and a range of 0.018.
  • Student B's results are 1.968 g, 1.969 g, and 1.973 g, with an average of 1.970 g and a range of 0.005.
  • Student C measures 2.001 g, 2.002 g, and 2.003 g, with an average of 2.002 g and a range of 0.002.
  • Student C is the most accurate and precise.

Stoichiometry

• Stoichiometry deals with the quantities of materials consumed (reactants) and produced (products) in chemical reactions.

Relative Atomic Masses

• Relative atomic masses are measured because atoms are very small and cannot be weighed individually.
• Atom mass is determined relative to another experimentally.
• Carbon 12 is assigned a mass of exactly 12 atomic mass units and serves as the standard for measuring the atomic mass of other elements.

Average Atomic Mass

• The atomic mass of carbon is 12.01 instead of 12.00 as a result of isotopes with the same atomic number and different mass numbers.
• Average atomic mass can be calculated using the relative abundance of isotopes.
• For example:
  • 12C has a relative abundance of 98.89%.
  • 13C has a relative abundance of 1.11%.
• The average atomic mass is calculated by ((98.89 / 100) x 12.00) + ((1.11 / 100) x 13.00) = 12.01.
• For example:
  • The atomic masses of 35/17CL (75.53 %) and 37/17 CL(24.47 %) are 34.968 amu and 36.956 amu respectively.
• To determine the average atomic mass of chlorine:
• (75.53/100 * 34.968) + (24.47/100 * 36.956)
• = 26.41 + 9.043 = 35.45

Avogadro's number and the Mole

• A mole is the substance of an amount that contains as many atoms, molecules, or other particles as there are in exactly 12 g of the carbon-12 isotope.
• A unit of measure is called the mole has been established for use in counting atoms due to large sample sizes.
• The mole represents the number equal to the number of carbon atoms in exactly 12 g of pure 12C.
• 12 g of 12C contains 6.023 x 1023 atoms.
• 6.023 x 1023 is Avogadro's Number.
• 1 mole of 12C= 6.023 x 1023 atoms
• 1 mole of H2O molecules = 6.023 x 1023 H2O molecules
• 1 mole of NO3- ions = 6.023 x 1023 NO3- ions
• 1 mole of H atoms = 6.023 x 1023 atoms
• 1 mole of O atoms = 6.023 x 1023 atoms
• 1 mole of H2 molecules = 6.023 x 1023 molecules
• To calculate the number of H atoms in 0.350 mol of C6H12O6:
• 1 mole of C6H12O6 = 12 moles of H atoms
• 0.350 mole of C6H12O6 → (0.350 / 1) x 12 moles of H atoms
• 4.2 moles of H atom → 6.023 x 1023
• Answer = 2.53 x 1024

Molar Mass

• The molar mass of a substance is the mass (in grams) of one mole of a compound.
• This is obtained by summing all combined atoms in one substance.
• Ex) 1 mole of Mg atom is 24.31 g therefore, the molar mass is 24.31 g/mol
  • The molar mass of methane (CH4) is CH4 = (12.01 x 1) + (1.008 X 4)= 16.04 g/mol

Moles and Molar Mass

• Number of moles = mass / Molar mass
• Percentage composition is used to determine elements in a compound. This is done by molar mass and determining mass ratios and percent.