Amount Of Substance Flashcards
One mole contains 6.02 x 10^23 atoms - what is this known as ?
Avagadros constant
What is the equation for number of particles?
Number of moles x Avagadros constant
What is the equation for number of moles?
Mass of substance / RFM
concentration (mol dm-3) x volume (dm3)
Ideal gas equation ?
pV = nRT
What do each variable stand for in pV = nRT ?
P - pressure (Pa)
V - volume (m^3)
n - number of moles
R - 8.31 J K-1mol-1 (always given)
T - temperature (K)
At a temperature of 60 degrees and a pressure of 250 kPa, a gas occupied a volume of 1100cm^3 and had a mass of 1.6g.
find its relative molecular mass
the gas constant is 8.31 J K-1mol-1
p = 250 x 1000 = 250 x10^3 Pa
V = 1100 x (1x10^-6) = 1.1 x 10^3 m^3
T = 60 +273 = 333 K
n = pV/RT
(250 x 10^3) x (1.1 x 10^3) / 8.21 x 333 = 0.0994 moles
Mr of gas = mass / number of moles = 1.6 g / 0.0994 = 16.1
Balance the equation
C2H6 + O2 –> CO2 + H2O
C2H6 + 3.5 O2 –> 2 CO2 + 3H2O
ionic equations only show ______ particles
reacting
example
HNO3 + NaOH –> NaNO3 + H2O
break up into ions
H+ + NO3- + Na+ + OH- –> Na+ + NO3- +H2O
cancel out
H+ + OH- –> H2O
Balanced equations can be used to work out masses
calculate mass of iron oxide produced if 27.9g pf iron is burnt in air.
4Fe + 3O2 –> 2Fe2O3
Mr of Fe = 55.8
mass/ RFM
27.9/55.8 = 0.5 moles
4 moles of Fe gives 2 moles of Fe2O3 so 0.5 moles of Fe would give 0.25 moles of Fe2O3
Mr of Fe2O3 =(2 x 55.8) + ( 3 x 16.0) = 159.6
moles x Mr = 0.25 x 159.6 = 39.9g
How to make a standard solution for titration ?
- Measure out solid on a balance, in a weighing boat
- Add into a beaker and wash weighing boat with distilled water, and add distilled water and stir with a glass rod to make sure the solid dissolves
- Pour into a volumetric flask, using a funnel
- Rinse beaker to make sure nothing is left behind
- Fill to 250 ml / to line with more distilled water until the bottom of the meniscus touches the line
- Invert to mix
titration method?
- Do a rough titration to get an idea of the end point. Add the acid to alkali using a burette, giving the flask a regular swirl
- Take an initial reading o how much acid is in the burette. \run the acid in to within 2cm3 of the end point. When you get to this stage add it in dropwise and keep swirling until the colour changes
- Work out the amount of acid used to neutralise the alkali. Final - initial, this volume is known as the titre
- Repeat the titration a few times until you have concordant results
- Use the results from each concordant result to calculate the mean volume of acid used.
indicators for titration
Methyl orange turns _____
phenolphthalein turns _____
red in acid –> yellow in alkali
colourless in acid –> pink in alkali
In a titration experiment, 25cm3 of 0.5 mol dm-3 HCl neutralised 35cm3 of NaOH solution
Calculate the concentration of the sodium hydroxide solution in mol dm-3
first write. balanced equation and write what you have
HCl + NaOH –> NaCl + H2O
25cm3 35 cm3
0.5 ??
moldm-3
mol of HCl 0.5 x 25/1000 = 0.0125 moles
from the equation we know that 1 mole of HCl neutralises 1 mole of NaOH
so 0.0125 moles of HCl. must neutralise 0.0125 moles of NaOH
conc of NaOH = moles / vol (dm3) = 0.0125 / (35/1000) = 0.357 mol dm-3
Define empirical formula
simplest whole number ratio of atoms of each element in a compound
Define molecular formula
actual number of atoms of each element in a compound
A molecule with Mr = 166 has the empirical formula of C4H3O2
find molecular formula
(4 x 12) + (3 x 1) + (2 x 16) = 48 + 3 + 32 = 83
166/83
= 2 empirical units in the molecule
molecular formula must be empirical x 2 so C8H6O4
A compound is found to have percentage composition
56.5% potassium
8.7% carbon
34.8% oxygen
find its empirical formula
in 100g there would be
mass/ Mr
56.5/39.1 = 1.445 mol of K
8.7/12 = 0.725 mol of C
34.8/16 = 2.175 mol of O
divide each number by the smallest
K = 2.00
C = 1.00
O = 3.01
ratio is 2 : 1 : 3
K2CO3
Percentage yield = _______________ / _____________ x ___
actual yield / theoretical yield x 100
0.475 g of CH3Br reacts with excess NaOH in the following reaction
CH3Br + NaOH –> CH3OH + NaBr
0.153 g of CH3OH is produced
What is the percentage yield?
number of moles of CH3Br = mass/ Mr = 0.475 / (12 + ( 3 x 1) + 79.9 ) +0.00501 mol
from the equation, moles of CH3OH is 1:1, so 0.00501 moles of CH3OH should form
Mr of CH3OH = 32
Theoretical yield = 0.00501 x 32 = 0.160 g
so
0.153 / 0.160 x 100 = 95.6%
% atom economy = ???
molecular mass of desired product / sum of molecular masses of all reactants x 100
Ethanol ( C2H12OH) can be produced by fermenting glucose (C6H12O6)
C6H12O6 –> 2C2H5OH + 2CO2
calculate atom economy
Always make sure the equation is balanced
2 x ((2 x 12 ) + ( 5 x 1) + ( 16 + 1 )) / ( 6 x 12) + ( 12 x 1 ) + (6 x 16) x 100 =
92/180 x 100 = 51.1%
companies want a high atom economy - less waste, less expensive, more sustainable
