3 2 the mole and the avogadro constant

The total number of ions in one mole of NaCl is: 2 x 6.02 x 1023 = 1.204 x 1024

True. There are 6.02 x 1023 carbon dioxide molecules in one mole of carbon dioxide.

A mole is the SI unit of amount of substance.

The Avogadro Constant is 6.02 x 1023, which is the number of particles in one mole of a substance.

True. One mole of any substance contains the same number of particles.

Molar mass is the mass of 1 mole of a substance.

RTP is room temperature and pressure, specifically 20°C and 1 atmosphere.

The molar gas volume at RTP is 24 dm3 or 24,000 cm3.

True. The molar volume of a gas is independent of the type of gas.

The equation for calculating the volume of a gas at RTP is: Volume = Moles x Molar Volume

4 moles of hydrogen gas contains:

To convert from cm3 to dm3, you divide by 1000.

The equation for molar gas volume using moles and volume is: Molar gas volume = volume / moles

The equation to calculate moles using volume and molar gas volume is: Moles = volume / molar gas volume

The amount of space, in dm3, that 1.5 moles of gas occupies is:

True. To convert volume from dm3 to cm3, you multiply by 1000.

A 300 dm3 cylinder contains 300 000 cm3 of gas.

To calculate the number of moles:

To calculate the mass:

To calculate the molar mass:

One mole of any element is equal to the relative atomic mass of that element in grams.

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol).

The equation for calculating moles from mass and Mr is: Moles = mass / Mr

False. The mass of 1 mole of a compound in grams is equal to its Mr.

The molar mass of carbon dioxide, CO2, is 44 g/mol.

The number of moles in 2.64 g of sucrose is:

The number of atoms in 15.7 g of water is: One mole of water contains: So, 0.872 moles of water contains:

False. When completing reacting masses calculations, the mass unit (e.g. grams, tonnes) does not affect the calculation. The mass unit only affects the final anwer as it must be given in the appropriate units.

The correct order of steps to complete a reacting mass calculation is:

CuCO3 CuO + CO2 The molar ratio of copper carbonate to copper oxide is 1 : 1.

2Na + Cl2 2NaCl The molar ratio of chlorine to sodium chloride is 1 : 2.

False. The masses of reactants and products in a reaction can be used to write a balanced equation for the reaction.

How the masses of the reactants and products be used to form a balanced chemical equation:

False. 3A + B 2C + 2D The molar ratio of A : C is not 2 : 3. There are 3 moles of A and 2 moles of C, which means the ratio is 3 : 2.

The moles, mass, molar mass equation that is needed for reacting mass questions is: moles = mass / molar mass

A limiting reactant is the reactant that is completely consumed in a chemical reaction and limits the amount of product formed.

False. The limiting reactant is the reactant that produces the least amount of product according to the balanced equation and the given quantities.

The first step in determining the limiting reactant is to convert the mass of each reactant into moles.

A reactant is in excess when there is still some of that reactant left after a chemical reaction is complete.

6.0 g of magnesium = 6.0 / 24 = 0.25 moles 2Mg + O2 2MgO The ratio of Mg : MgO is 1 : 1 So, 0.25 moles of magnesium oxide can be produced from 6.0 g of magnesium.

True. In a reaction, the limiting reactant determines the maximum amount of product that can be formed.

Mr (Al2O3) = 102. Moles of aluminium oxide = 51 / 102 = 0.5. The ratio of Al2O3 : Al is 1 : 2. So, 0.5 moles of aluminium oxide will produce 1.0 moles of aluminium. So, the maximum mass of aluminum that can be produced is 27 tonnes.

A chemical equation should be balanced before performing reacting mass calculations to ensure that the correct molar ratios are used in the calculations.

Concentration refers to the amount of solid / solute there is in a specific volume of the liquid / solvent.

Using moles, volume in dm3 and concentration, the equation is: Moles = concentration x volume

To calculate the number of moles of solute present in 2.0 dm3 of a solution whose concentration is 0.15 mol dm-3:

The two common units for concentration are:

To convert g dm-3 to mol dm-3, you divide the concentration in g dm-3 by the molar mass.

The equation for concentration using moles and volume is: concentration = moles / volume

The concentration of a solution where 1.0 mole of solute is dissolved in 2.0 dm3 of water is: Concentration = moles / volume Concentration = 1.0 / 2.0 = 0.5 mol dm-3

False. To convert dm3 into cm3, you multiply by 1000.

The equation to calculate the volume of a solution using moles and concentration is: Volume = moles / concentration

Using moles, volume in dm3 and concentration, the equation is: Moles = concentration x volume

To calculate the number of moles present in 2.5 dm3 of a solution which has a concentration of 0.30 mol dm-3:

The balanced equation is: H2SO4 + 2KOH → K2SO4 + 2H2O

27.50 cm3 is 0.0275 dm3.

The ratio of acid to alkali is 1:3. H3PO4 + 3NaOH → Na2PO4 + 3H2O

The concentration in mol/dm3 of a solution of HCl with a volume of 23.55 cm3 and contains 0.00375 moles is: Volume in dm3 = 0.02355 dm3 Concentration = = 0.16 mol/dm3

The four steps in a titration calculation are:

Using moles, volume in dm3 the equation is: Concentration =

0.03905 dm3 is 39.05 cm3.

0.05 moles of HCl would be required as this reaction is a 1:1 ratio. HCl + NaOH → NaCl + H2O

The molecular formula shows the actual number of atoms of each element in a molecule, while the empirical formula shows the simplest ratio.

The steps to calculate an empirical formula are:

The molecular formula gives the actual numbers of atoms of each element present in one molecule of a compound.

To calculate the molecular formula from the empirical formula, multiply the subscripts in the empirical formula by n, where n = (molecular mass / empirical formula mass).

True. If the molecular formula of benzene is C6H6 the empirical formula of benzene is CH.

A hydrated salt is a crystallised salt that contains water molecules as part of its structure.

True. The empirical formula is CH.

The mistake in the empirical formula calculation is that the mole calculations are incorrect / upside down. The correct calculation is: The empirical formula is correctly calculated to be MgCl2.

The mistake in the empirical formula calculation is that the atomic number has been used, not the atomic mass. The correct calculation is: So, the empirical formula is CH3.

The mass of the empirical formula, CH3, is 12 + (1 x 3) = 15. 30 / 15 = 2, which means that two lots of the empirical formula are needed. So, the molecular formula is C2H6.

Percentage yield compares the actual yield to the theoretical yield.

Actual yield is the recorded amount of product obtained from a chemical reaction.

The actual yield can be determined by experiment only.

Theoretical yield is the amount of product that would be obtained under perfect practical and chemical conditions.

The type of calculation lets you determine theoretical yield is a reacting mass calculation.

The equation for percentage yield is: (actual yield theoretical yield) x 100

True. For economic reasons, the objective is to have as high a percentage yield as possible to increase profits and reduce costs and waste.

If the actual yield is 1.50 g and theoretical yield is 2.00g, the percentage yield is: (1.50 2.0) x 100 = 75%

False. A percentage yield of 100% is not common in chemical reactions.

Factors that can reduce percentage yield include:

Percentage composition by mass is the mass of each element expressed as a percentage of the total mass of the compound.

The equation to calulcate percentage composition by mass is: (total mass of the element in the compound / relative formula mass of the compound) × 100

True. The sum of percentage compositions by mass for all elements in a compound should equal 100%.

False. So, the percentage by mass of Mg in MgO is 24 / 40 x 100 = 60%.

True. The mass of nitrogen in ammonium nitrate, NH4NO3, is 28 because there are two nitrogen atoms in the compound.

The percentage of fluorine in tin(II) fluoride, SnF2 is:

Percentage purity is the mass of pure substance expressed as a percentage of the total mass of the impure substance.

The equation for percentage purity is: (mass of pure substance / total mass of substance) × 100

True. A sample with 100% purity contains no impurities.

The percentage purity of the calcium chloride is: (9.8 / 12.0) x 100 = 81.7%