VCE+U1+3.+Moles+and+Masses

=Chap 4. Summary Moles and Masses= In Chemistry we measure quantities in 2 main ways. We measure a chemical in grams by determining its molecular mass. We measure a chemical in moles also.

The big numbers in a chemical equation (the ones in front of a compound) represents the number of moles required for the reaction to go to completion.

We can convert the quantities of moles to grams and vice versa. Use this relationship number of moles = the mass I have / the molar mass

n = m/M

When we use Avagadro's number we are measuring the number of particles in a mole. number moles x 6.02 x10 23 = number of particles (I have or in the question) n = number of particles I have / number of particles in a mole ( ie Avagadro's number)

n = N / **Na**

The **Relative Molecular Mass** **Simulation -** http://phet.colorado.edu/en/simulation/isotopes-and-atomic-mass

(RMM, symbol, **Mr**) is the total of the Relative Atomic Mass of each atom in a compound. There is no unit for Mr. If a dot (.) is in the formula, the atomic masses are added not multiplied. eg. Mr (CuSO4.5H2O) = 63.5 (Cu) + 32 (S) + 4 x 16 (O) + 5 x [2 x 1 (H) + 16 (O)] = 249.5.

For ionic compounds that are made up of many units of the cation and anion, the Molecular Mass is called the **Formula Mass** and is equal to the Mr of the ionic compound’s formula. eg. Mr(NaCl) = 23 + 35.5 = 58.8

The **Molar Mass** (symbol, **M**) is the mass of 1 mole of any substance and is numerically equal to the Relative Molecular Mass but has a unit _. Therefore the molar mass of CuSO4.5H2O = 249.5 g mol-1.

The **Mole** (symbol, n and unit, mol) is the **Amount Of Substance** containing a fixed number of particles and is the standard way Chemists describe the amount of substance in different states (solid, solution, liquid or gas). This is not a concentration. Concentration is the number of moles per litre - and we'll deal with that when we study acids and bases.
 * Measuring the Amount of Different Substances: **

The number of particles in 1 mole of any substance is equal to the value of **Avogadro's Constant** (symbol **Na**) which is **6.02 x 10** 23 **mol-1**. This value is also equal to the number of particles in _ of Carbon. To use the different mol formulae correctly, the correct units for each substance’s measure must be used:- Mass, **m** is in grams, **//g//**. (1000 mg = 1.0 g, 1.0 kg = 1000 g and 1 tonne = 1000 kg). Volume, **V** is in Litres, **//L//**. (1000 mL (cm 3 ) = 1.0 L (dm 3 ) and 1 m 3 = 1000 dm 3 ). Concentration, **c** is in Molarity, **//M//** or **//mol/L//**. Pressure, **P** is in kiloPascals, **//kPa//**. (101.325 kPa = 101325 Pa = 760 mmHg = 1.0 atmosphere pressure). Temperature, **T** is on Kelvin, **//K//** (Temperature in Kelvin = Temperature in degrees Celsius, oC + 273 K). For particles (atoms, molecules, ions), use, the amount, **n** in **mol =** For masses of solids, liquids or gases, use, the amount, **n** in **mol =** For _ (not pure liquids), use, the amount, **n** in **mol = Concentration, c in M x Volume, V** in **L Covered later** For gases, use, the amount, **n** in **mol =** covered later

In calculations, the final answer is written with the same number of significant figures as the value with the __least number of significant figures__ used in the calculation. Any 0 before any digit (including a decimal point) is not a significant figure but all digits including & after the first non-zero digit are counted as significant figures. eg. 0.12 has 2 sig. figures, 1.02 has 3 sig. figures, 20.00 has 4 sig. figures and 0.002 has 1 sig. figure (2 x 10-3). If values are very low or high use Standard Form. eg. 30257 g = 3.0 x 104 g and for 0.003 g = 3.0 x 10-3 g. 1.20 x 1024 molecules of H2O. // Number of particles = 1.20 x 1024, Find n //
 * When doing mol calculations, ensure the Chemical is shown in a after writing n. eg. n(Chemical) = **
 * Basic question to answer when doing calcuations - “How many moles is that” **
 * ==== Determining the Amount in Mol ==== || ==== Determining another unit using the Amount in mol ==== ||
 * ==== For Particles ==== || ==== For Particles ==== ||
 * 1. Determine the **amount (in mol)** present in

n(H2O)= 1.20 x 10 24 / 6.02 x 10 23

n(H2O)= 2.00 mol || 1. Determine the **number of atoms** in 2.0 mol of H2O. Number of particles = n x Avogadros Constant. // n = 2.0, Find the number of particles //

Number of H2O= n(H2O) x 6.02 x 1023. = 2.00 x 6.02 x 1023 = 1.20 x 1024 H2O molecules. 1 H2O molecule has 2 x H + 1 x O = 3 atoms Number of atoms = 3 x 1.2 x 1024 = 3.6 x 1024 atoms || 12 g of Oxygen. // m = 12 g, Find n // // Oxygen like most non metals (not Noble gases) exists as diatomic molecules (2 atoms) = O2. // n(O2) = = n(O2) = 0.38 mol (not 0.375) due to sig figs || 2. Determine the **mass of Magnesium** present in 3.3 mol of Magnesium. // n = 3.3, Find the mass // // Magnesium like all metals, exists as single atoms = Mg. //
 * ==== For Masses of solid/liquid/gas solutes ==== || ==== For Masses of solid/liquid/gas solutes ==== ||
 * 2. Determine the **amount (in mol)** present in

m(Mg) = n(Mg) x Mr(Mg), The Mr are on Pg 313. = __ m(Mg) = __ || present in 500 mL of 0.25 M NaCl solution.
 * ==== For solutes in Solutions ==== || ==== For solutes in Solutions ==== ||
 * 3. Determine the **amount (in mol)** of NaCl

V = 500 mL = 0.500 L (mL ÷ 1000 = L), c = 0.25 M, Find n
n(NaCl) = Conc. c in M x Volume, V in L = __ n(NaCl) = __ || 3. Determine the **concentration of NaCl** present in a 2.0 L solution containing 0.12 mol of NaCl. // n = 0.12, V = 2.0 L, Find c // [NaCl] = The [ ] represents the conc. = [NaCl] = 0.060 M. || in 2.0 L of 0.25 M Glucose solution
 * Some calculations involve the combination of 2 or more Mol formula to determine a measurement. Use the measurements stated to determine the amount in mol (n) and then use this to find the required measurement. ||
 * Determine the **mass of Glucose**, C6H12O6 present

V = 2.0 L, c = 0.25 M, Find n and then find m
n(C6H12O6) = c x V = 0.25 x 2.0 = 0.50 mol m(C6H12O6) = n(C6H12O6) x Mr = 0.50 x 180 m(C6H12O6) = 90 g. || Determine the **number of Glucose**, C6H12O6 **molecules** present in 2.0 L of 0.50 M Glucose solution V = 2.0 L, c = 0.25 M, Find n and then find number of molecules n(C6H12O6) = c x V = 0.50 x 2.0 = 1.0 mol No. of C6H12O6= n(C6H12O6) x Na (6.02 x 1023) = 1 x 6.02 x 1023 = 6.0 x 1023 C6H12O6 molecules ||

=Empirical formula calculations= Facts: an empirical formula is like the lowest common denominator. Eg CH 3 is the empirical formula C 2 H 6 is not - it is a molecular formula and as such has a specific mass. (in this case 30 g 2 x 12 + 1 x 6) C 3 H 9 is also a molecular formula and has a mass of 45g (12 x 3 +1 x 9)

The other thing the empirical formula tells us is the ratio in which the atoms of the elements combine. For our example above CH 3 it is 1 : 3, (C : H). It also tells us the atoms of the elements combine in the molar ratio of 1:3.

The molecular formula is dependent on the compounds molecular mass. For example if a subtance has a molecular mass of 45g and a formula mass ( based on its empirical formula) of 15g then we can predict that there are 3 empirical units in this molecule. and hence the molecular formula will be 3 x CH 3 = C 3 H 9

Example
A compound contains 27.3% C and 72.7% oxygen. Calculate the empirical formula. Record the mass in grams. (m) If % is given assume you have 100g || 27.3 || 72.7 || Calculate the amount in moles (n) || 27.3/12.0 =2.27 || 72.7 / 16.0 =4.54 || Divide each of the moles by the smallest mole value in step 2 || 2.27 /2.27 =1 || 4.54 /2.27 =2 || Obtain the simplest whole number mole ratio || 1 || 2 || therefore the empirical formula CO 2
 * || Carbon || Oxygen ||
 * Step 1.
 * Step 2.
 * Step 3.
 * Step 4

Example 2
A sample of a compound of aluminium and oxygen has a mass of 2.36g. It contains 1.25g Al. Find the empirical formula

0.0463 || 1.11 / 16.0 = 0.0694 || 1.0 || 0.0694 / 0.0463 = 1.5 ||
 * ||  || Al || O ||
 * Step 1 || m(g) || 1.25 || 2.36-1.25=1.11 ||
 * Step 2 || n (mol) || 1.25 / 27.0=
 * Step 3 || Divide all amounts by the smallest number of moles || 0.0463 /0.0463 =
 * Step 4 || Multiply both numbers by a factor that makes all the numbers in step 3 whole numbers ( in this case 2) || 2 || 3 ||

Therefore the empirical formula is Al 2 O 3

= Calculating Molecular formula = Overview - calculate the empirical formula, then use the mass given to identify the number of moles

Example
Hydrocarbon contains 7.2g of C and 1.5g of H. The molar mass of the compound was 58 g mol -1. = 0.6 || 1.5 / 1.0 = 1.5 || = 1 || 1.5 / 0.6 = 2.5 || So the empirical formula is C 2 H 5 the molecular mass of C 2 H 5 is 12 x 2 + 1 x 5 = 29g But the molar mass of the compound is 58g ( data given from the question) Number of C 2 H 5 s in one of these 58g moles is 58 / 29 = 2 therefore the molecuar formula is 2 x the empirical formula ie C 4 H 10
 * ||  || C || H ||
 * Step 1 || m (g) || 7.2 || 1.5 ||
 * Step 2 || n (mol) || 7.2 / 12
 * Step 3 || Divide all amounts by the smallest number of moles from Step 2 || 0.6 / 0.6
 * Step 4 || Give simplest whole number mole ratio (x 2) || 2 || 5 ||

= Extension activity =

http://phet.colorado.edu/en/simulation/isotopes-and-atomic-mass

http://phet.colorado.edu/en/simulation/molarity