Chemistry AS - Chapter 02 - Amount of Substance

Revision cards for Chapter 2 - Amount of Substance.

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Relative Atomic and Molecular Masses, The Avogadro

We compare the masses of atoms by using relative masses.

Scientists use carbon 12 as the baseline because the mass spectrometer allowed us to measure the masses of individual isotopes so accurately, 1/12th of Carbon 12 is exactly 1.

The R.A.M is the weighted average mass of an atom of an element, taking into account its naturally occuring isotopes, relative to 1/12th of a Carbon 12 atom.

The relative molecular mass (R.M.M) is the mass of a molecule compared to that of an atom of carbon 12.

Atomic Mass (Ar) = (Average mass of 1 atom/mass of Carbon 12) x 12

Molecular Mass (Mr) = (Average mass of 1 molecule/mass of Carbon 12) x 12

One atom of any element is too small to see and impossible to weigh.

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Relative Atomic and Molecular Masses, The Avogadro

Avogadro constant is the number of atoms in 12 grams of Carbon 12.

A mole contains 6.022 x 10^23 particles.

The R.A.M of any element in grams contains one mole of atoms.

The R.M.M of a substance in grams contains one mole of entities.

We can also have moles of ions and electons.

Number of moles = mass(g)/mass of 1 mole

"Mass of 1 mole" means R.A.M or R.M.M

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Ideal Gas Equation

Ideal gas equation is:

PV=nRT

R is the gas constant. R is 8.31Jk^-1mol^-1

No gases obey this equation but at room temperature and pressure, it holds quite well for many gases.

Units:

• P must be in Pa (Pascals) or Nm^-2 (Newton metres squared)
• V must be in m^3 (metres cubed)
• T must be in K (Kelvin)
• R must be in JK^-1mol^-1

We can rearrange this equation to find volume, temperature, pressure and no. of moles.

If we know the number of moles present in a given mass of gas, we can find the mass of 1 mole of gas. This would tell us the R.M.M

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Empirical Formulae and Molecular Formulae

Empirical Formula is the formula that represents the simplest ratio of atoms present in a compound.

To find the empirical formula:

• Find the masses of each of the elements present.
• Work out the number of moles of atoms in each element.
• Convert the no. of moles of each element into a whole number ratio.

The molecular formula gives the actual number of atoms of each element in one molecule of the compound.

Empirical Formula is not always the same as the molecular formula.

To find the number of units of the empirical formula in the molecular formula, divide the R.M.M by the relative mass of the empirical formula.

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Moles in Solutions

The concentration of solution tells us how much solute is present in a known volume of solution.

Measured in mol dm^-3 (moles per dm cubed)

1 mol dm^-3 means there is 1 mole of solute per cubic decimetre of solution.

Concentration = No. of moles/volume(dm^3)

Number of moles in a solution = (mxv)/1000

Combine the equations and you get:

Number of moles in a solution = (volume/1000)x concentration

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Balanced Equations and Related Calculations

Balanced equations means both sides of the equation have the same number of molecules of each element. Equations should have state symbols as well.

In some reactions, we simplify the equation by considering the ions present. These are ionic equations.

Sometimes there are ions that do not take place in the reaction.

HCl + NaOH ------> NaCl +H2O

The ions present are:

• HCl ----> H+ and Cl-
• NaOH --> Na+ and OH-
• NaCl ----> Na+ and Cl -

If you write the equation and strike out ions that appear on each side, you will end up with:

H+ + OH-  ------> H2O (Na+ and Cl- are spectator ions.)

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Balanced Equations and Related Calculations

Tips for balancing equations:

• Use correct formulae.
• Only change numbers of atoms by putting a number (coefficient) in front of formulae.
• The coefficient tells you how many moles of a substance you are reacting.
• With ionic equations, the charge on each side must be the same.
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Balanced Equations, Atom Economies and Percentage

Atom economy tells us in theory how many atoms must be wasted in a reaction.

Atom Economy = (mass of desired product/total mass of reactants) x 100

Yield tells us the practical efficiency of the process.

Yield of a chemical reaction =

(number of moles of specified product/theoretical max. number of grams of product) x 100

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