Transcript Energy Transfer - Noadswood School

Yield
Noadswood Science, 2012
Thursday, July 16, 2015
Yield

To be able to calculate the yield from chemical reactions
Conservation Of Mass

What does conservation of mass mean?

Mass is never lost or gained in chemical reactions

Mass is always conserved – the total mass of products at the end of the
reaction is equal to the total mass of the reactants at the beginning

This fact allows you to work out the mass of one substance in a reaction if
the masses of the other substances are known…
Practice

Often in chemical reactions is appears that mass has been lost / gained –
why could this be?

In practice, it is not always possible to get all of the calculated amount of
product from a reaction:  Reversible reactions may not go to completion
 Some product may be lost when it is removed from the reaction mixture
 Some of the reactants may react in an unexpected way
 The reactants could react with something which was not measured (i.e.
oxygen within the air would add mass to the final product)
 Some of the products might be hard to measure (i.e. a gas could be given
off from the reaction)
Reacting Masses

To calculate the mass in reactions there are three steps:  Write out the balanced equation
 Work out the Mr (off the bits you want)
 Apply the rule ‘divide to get one, multiply to get all (first to the
substance given, then to the one with no information)’!

E.g. what mass of magnesium oxide is produced when 60g of magnesium is
burned in the air?
Reacting Masses

E.g. what mass of magnesium oxide is produced when 60g of magnesium is burned in the air?
2Mg + O2 → 2MgO

Relative formula: (2 x 24) → 2 x (24 + 16)
48

80
Apply the rule: divide to get one, multiply to get all
48g Mg reacts to give 80g MgO
÷ 48
÷ 48
1g Mg reacts to give 1.67g MgO
x 60
x 60
60g Mg reacts to give 100g MgO
Reacting Masses

To calculate the mass in reactions there are three steps:  Write out the balanced equation
 Work out the Mr (off the bits you want)
 Apply the rule ‘divide to get one, multiply to get all (first to the
substance given, then to the one with no information)’!

E.g. if we have 50g of CaCO3, how much CaO can we make?
Reacting Masses

E.g. if we have 50g of CaCO3, how much CaO can we make?
CaCO3 → CaO + CO2

Relative formula: 40 + 12 (3 x 16) → 40 + 16
100

56
Apply the rule: divide to get one, multiply to get all
÷ 100
100g CaCO3 reacts to give 56g CaO
÷ 100
1g CaCO3reacts to give 0.56g CaO
x 50
x 50
50g CaCO3reacts to give 28g CaO
Reacting Masses

To calculate the mass in reactions there are three steps:  Write out the balanced equation
 Work out the Mr (off the bits you want)
 Apply the rule ‘divide to get one, multiply to get all (first to the
substance given, then to the one with no information)’!

E.g. what mass of carbon will react with hydrogen to produce 24.6g of
propane?
Reacting Masses

E.g. what mass of carbon will react with hydrogen to produce 24.6g of propane?
3C + 4H2 → C3H8

Relative formula: 3 x 12 → (3 x 12) + (8 x 1)
36

44
Apply the rule: divide to get one, multiply to get all
÷ 44
36g C reacts to give 44g C3H8
÷ 44
0.82g C reacts to give 1g C3H8
x 24.6
x 24.6
20.1g C reacts to give 24.6g C3H8
Yield

The amount of product made is called the yield – in a chemical reaction no
atoms are lost or gained but sometimes the yield is not what you would
expect

Theoretical yield: maximum products that are made if reactants react

Actual yield: the amount of product which actually forms
Percentage Yield

The yield of a reaction is the actual mass of product obtained – the
percentage yield can be calculated: Percentage yield = mass product obtained x 100
theoretical mass

For example, the maximum theoretical mass of product in a certain
reaction is 20g, but only 15g is actually obtained…
Percentage yield = 15⁄20 × 100
= 75%
Experiment

Precipitation is the formation of an insoluble solid when two solutions are
mixed - e.g. barium sulfate is produced by precipitation from barium nitrate
and sodium sulfate solutions

Write a word and symbol equation for this reaction…
barium nitrate + sodium sulfate  barium sulfate + sodium nitrate
Ba(NO3)2 + Na2SO4  BaSO4 + 2NaNO3

Using the standard procedure carry out the precipitation reaction…
Experiment









Pour 50cm3 water into a 100cm3 beaker
Weigh 2.6g barium nitrate
Combine the two and stir (until all barium nitrate is dissolved)
Pour this into the 250cm3 beaker
Measure out 75cm3 sodium sulfate into a 100cm3 beaker
Add the two solutions together
Stir well (notice the white precipitate)
Filter the mixture using a funnel and filter paper (make sure you weigh your
filter paper) - wash the residue with a little water
Dry the precipitate - weigh to find our yield (- filter paper)!…
Calculating Yield

Calculate the maximum theoretical yield of barium sulfate and then work
out your own percentage yield for the experiment…

Ar of barium = 137; sulfur = 32; nitrogen = 14; and oxygen = 16
barium nitrate + sodium sulfate  barium sulfate + sodium nitrate
Ba(NO3)2 + Na2SO4  BaSO4 + 2NaNO3
Theoretical Yield
Ba(NO3)2 + Na2SO4  BaSO4 + 2NaNO3

Relative formula (barium nitrate and barium sulfate): 137 + (2 x 14) + (2 x (3 x 16)) → 137 + 32 + (4 x 16)
261

233
Apply the rule: divide to get one, multiply to get all
261g Ba(NO3)2 reacts to give 233g BaSO4
2.61g Ba(NO3)2 reacts to give 2.33g BaSO4
Maximum theoretical yield for experiment = 2.33g BaSO4
Percentage Yield

The yield of a reaction is the actual mass of product obtained – the
percentage yield can be calculated: Percentage yield = mass product obtained x 100
theoretical mass

In this experiment the maximum theoretical yield is 2.33g – if you got, for
example, 1.25g then
Percentage yield = 1.25⁄2.33 × 100
= 53.6%
Empirical Formula

Empirical formula is a simple expression of the relative numbers of each
type of atom in it…

The following steps are used to calculate empirical formula:  List all the elements in the compound
 Write underneath them their experimental masses or percentages
 Divide each mass or percentage by the Ar for that particular element
 Turn the numbers until you get a ratio by multiplying / dividing them by
well chosen numbers
 Get the ratio in its simplest form…
Empirical Formula – Example

Find the empirical formula of the iron oxide produced when 44.8g of iron react with 19.2g of
oxygen (Ar iron = 56; oxygen = 16)
Iron (Fe)
Oxygen (O)
Experiment mass
44.8g
19.2g
÷ Ar for each
element
44.8/56 = 0.8
19.2/16 = 1.2
x 10
8
12
÷4
2
3
Simplest formula is 2 atoms of Fe to 3 atoms of O (Fe2O3)
Empirical Formula – Example

Find the empirical formula of sulfur oxide if 3.2g of sulfur reacts with oxygen to produce 6.4g
sulfur oxide (Ar sulfur = 32; oxygen = 16)

Conservation of mass tells us that the mass of oxygen equals the mass of sulfur oxide minus
the mass of sulfur – the mass of oxygen reacted = 6.4 - 3.2 = 3.2g
Sulfur (S)
Oxygen (O)
Experiment mass
3.2g
3.2g
÷ Ar for each
element
3.2/32 = 0.1
3.2/16 = 0.2
x 10
1
2
Simplest formula is 1 atoms of S to 2 atoms of O (SO2)
Moles

A mole is a number – 6.023 x 1023

When you get precisely this number of atoms of carbon-12 it weighs 12g

That number of atoms or molecules of any element or compound will
weigh exactly the same number of grams as the relative atomic mass of the
element or compound: 
Iron has an Ar of 56 – 1 mole of iron weighs 56g

Nitrogen has a Mr of 28 (2 x 14) – 1 mole of nitrogen weighs 28g

Carbon dioxide has a Mr of 44 (12 + 2 x 16) – 1 mole of carbon dioxide
weighs 44g
Moles

To work out the number of moles in a given mass: -
Number of moles = Mass (g) of element or compound
Mr of element or compound

How many moles are there in 42g of carbon?
42/12 = 3.5 moles