Ch 01 summary


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Unformatted text preview: CHAPTER 1 QUANTITATIVE CHEMISTRY (IB TOPIC 1) SUMMARY Introduction • 1 dm3 = 1 litre = 1 x 10-3m3 = 1 x 103 cm3 = 1000 ml • Amount of substance, n, is measured in moles (mol). • 1 mol of a chemical species contains the same number of particles as there are atoms in exactly 12 g of C-12 ( 12 C ) isotope. 6 • 1 mol of any substance contains 6.02 × 1023 particles; 6.02 × 1023 mol−1 is called Avogadro’s Constant (L or NA) • Ar, the relative atomic mass of an element is the ratio of the mass of an atom of the element to the mass of one atom of C-12; Ar has no units. • Mr, the relative molecular mass also has no units. • M, the molar mass is the mass of one mole of any chemical species and has the units g mol−1. The amount of substance, n in moles = n= mass ( g ) ; molar mass ( g mol −1 ) V (dm 3 ) . Vm (dm 3 mol −1 ) ⇔ Number of Particles Avogadro Constant (6.02×1023 mol−1) Amount ⇔ Mass in moles Molar mass in grams • Use Avogadro Constant to convert between amount and number of particles. • Use molar mass of substance to convert between amount in moles and mass in grams. • Empirical formula gives the simplest whole number ratio of atoms in a compound. • Molecular formula gives the actual number of atoms of each element in the molecule of a compound. If a molecular formula is given, percentage composition can be calculated. If the percentage composition is given: • Consider 100 g of sample, the % of each element becomes its mass. • Convert the mass of each element to its amount in moles. • Determine simplest whole number ratio – this is its empirical formula. • Molecular formula is a whole-number multiple of the empirical formula. • Substances react by amounts based on a balanced chemical equation. © IBID Press 2007 1 CHAPTER 1 QUANTITATIVE CHEMISTRY (IB TOPIC 1) SUMMARY If masses are given, convert these to amounts of substances. • In stoichiometric problems, apply mole ratios as specified in the balanced chemical equation. • The % yield indicates how efficient a reaction is (given by: experimental yield ×100% ). theoretical yield • STP for gases is standard temperature (0°C or 273 K) and pressure (1 atmosphere or 101.325 kPa). • Molar volume, Vm, of any gas at STP = 22.4 dm3. In a balanced chemical equation, coefficients stand for the amount of substance. For gases, these also refer to volumes of gases. Concentration, c = amount n (mol) n = mol dm −3 . 3 Vol solution (dm ) V n = cV; volume must be in dm3. On dilution, the amount of solute does not change but the volume increases and the concentration decreases Amount n = m (g) . M (g mol −1 ) For a reaction aA + bB → products, where a and b are coefficients, then 1 1 a nA = b nB . Percentage yield = experimental yield × 100% . theoretical yield (There are many worked examples given in the chapter) © IBID Press 2007 2 ...
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This document was uploaded on 10/18/2011.

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