SC9 Flashcards
SC9b
1) Explain the law of conservation of mass in a closed system.
2) Explain the law of conservation of mass in a non-enclosed system.
1) No substances can enter or leave a closed system. A simple closed system could be a sealed container such as a stoppered flask. Sometimes reactions that happen in open beakers are closed systems, for example if there are no gasses in the reaction, the reactants or products cannot escape. The total mass of the beaker and the substances it contains stays the same during the reaction.
2) Substances can enter or leave a non-enclosed system. These systems include open flasks, boiling tubes or crucibles that let gases enter or leave. If a gas escapes, the total mass will look as if it has decreased. If a gas is gained, the total mass will look as if it has increased. However, the total mass stays the same if the mass of the gas is included.
SC9b
1) Calculate the mass of product formed from a given mass of reactant, using a balanced equation.
2) Calculate the mass of a reactant needed to produce a given amount of product, using a balanced equation.
1) Write the balanced equation.
Calculate and write the relative formula masses under the substances and add the units in the question
Multiply the relative formula masses by the coefficients in the equation
Find the mass of product for 1 g of reactant
Scale up for the mass given in the question
2) Eg. Calculate the mass of chorine needed to make 53.4g of aluminium chloride
Write the balanced equation
2Al + 3Cl2 —> 2AlCl3
Calculate relative formula masses of the substances needed
Cl2 = 2 x 35.5 = 71
AlCl3 = 27 + (3x35.5) = 133.5
Calculate ratio of masses (multiply Mr values by the balancing numbers shown in the equation).
3 x 71 = 213g Cl2
2 x 133.5 = 267g AlCl3
Work out the mass for 1g of reactant or product
Eg. 213/ 267 makes 267/267
0.798g Cl2 makes 1g AlCl3
Scale up or down (from 1g to the mass you are given)
53.4 x 0.798 = 42.6g Cl2
53.4g AlCl3
SC9b
1) What is the law of conservation of mass?
2) Calculate the concentration of a solution in g dm–3
1) The law of conservation of mass states that no matter is lost or gained during a chemical reaction. Mass is always conserved, therefore the total mass of the reactants is equal to the total mass of the products. The overall mass of the substances does not change.
2) Concentration in g dm-3 = mass of solute in g / volume of solution in dm3
SC9c
1) Describe what is meant by a mole of particles.
2) Calculate the number of moles of particles in a given mass of a certain substance and vice versa.
3) Calculate the number of particles in a given number of moles or mass of a substance and vice versa
1) The mole is the unit for amount of substance. It is abbreviated to mol. 1 mol is the amount of substance that contains the same number of particles as there are atoms in 12.0 g of carbon-12.
2) Number of moles of substance = mass (g) / Ar (small r) or Mr (small r)
3) Mass (g) = Number of moles of substance * Ar (small r) or Mr (small r)
SC9c
1) Explain what a limiting reactant is
2) Work out the balanced equation for a reaction using the masses of reactants and/or products, which is given in the question
1) The mass of a product formed in a reaction is controlled by the mass of reactant that is not in excess.
The mass of product formed in a reaction depends upon the mass of the limiting reactant. This is because no more product can form when the limiting reactant is all used up.
2) First calculate the number of moles for each element/ compound. Then divide these two answers by the number of the smaller moles. Find the simplest whole number ratio (so that the ratio is not a decimal), and finally complete the equation by including the formula of the product on the other side of the arrow, and balance the equation the normal way.
SC9a
1) How do you calculate the relative formula mass of a substance from relative atomic masses?
2) Calculate the empirical formula of a compound from the masses of the elements it contains.
3) Explain the difference between an empirical formula and a molecular formula.
1) The relative formula mass (Mr (small r))of a substance is the sum of the relative atomic masses of all the atoms of ions in its formula. Relative atomic masses are given in the periodic table.
2) A. Find the number of moles of atoms by dividing the mass by the mass of 1 mole of atoms. This will give you the number of moles of atoms.
B. Divide by the smaller number to find the ratio. This will give you the ratio of moles. Example: 0.2/0.1 = 2 for Cu, 0.01/0.01 = 1 for O.
C. Use this ratio of moles to create the empirical formula. (Cu2O)
3) An empirical formula is the simplest whole number ratio of atoms or ions of each element in a substance.
The molecular formula represents the actual number of atoms of each element in one molecule.
SC9a
1) Deduce the empirical formula from a molecular formula.
2) Deduce the molecular formula for a compound from its empirical formula and its relative formula mass.
1) A. First find the relative formula mass by adding up the atomic numbers of all the elements in the empirical formula. Eg. 2 x H mass + 3 X C mass + 1 x O mass.
B. Then, divide the relative formula mass (Mr, small r) by the empirical formula mass. To find the molecular formula, multiply this answer by the small numbers in the empirical formula. Eg. The empirical formula is CH2O, and the molecular formula is C6H12O6
2) To deduce the molecular formula for a compound from its empirical formula and its relative formula mass, you need to follow these steps:
1. Find the molecular formula mass by multiplying the empirical formula mass with a whole number (n).
2. Divide the relative formula mass by the molecular formula mass to get the value of n.
3. Multiply the subscripts in the empirical formula by n to get the molecular formula.
For example, suppose the empirical formula of a compound is CH and its relative formula mass is 78.
a. The empirical formula mass of CH is 12 + 1 = 13.
b. To find the molecular formula mass, we need to multiply the empirical formula mass by n. Let’s assume the molecular formula is CₙHₘ. Then the molecular formula mass will be 12n + 1m.
78 = (12n + 1m) / n
78n = 12n² + nm
3. Solving the above equation gives n = 6 and m = 6.
4. The molecular formula of the compound is C₆H₆.
SC9a
Describe an experiment to determine the empirical formula for a compound.
- Measure and record the mass of an empty crucible with its lid.
- Put a length of magnesium ribbon into the crucible. The is crucible where you carry out the reaction.
- Measure and record the total mass of the crucible, its lid and contents.
- Place the crucible on a tripod with a pipe clay triangle. Strongly heat the crucible for several minutes using a Bunsen burner.
- When the magnesium has stopped glowing, turn off the Bunsen burner and allow the crucible to cool down.
- Repeat this step: Measure and record the total mass of the crucible, its lid and contents.
