1 PERCENT COMPOSITION, EMPIRICAL & MOLECULAR FORMULAS

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PERCENT COMPOSITION, EMPIRICAL & MOLECULAR

FORMULAS

PERCENT COMPOSITION, EMPIRICAL & MOLECULAR

FORMULAS

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o New food labels are required to describe the ingredients using percents of the daily reccom- mended allowance

•These numbers tell what part of the total # of calories can be obtained

from a product

•AKA percent composition

o New food labels are required to describe the ingredients using percents of the daily reccom- mended allowance

•These numbers tell what part of the total # of calories can be obtained

from a product

•AKA percent composition

Percent CompositionPercent Composition

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o To get the information found on food labels the chemists had to know what fraction of the whole was each component

•Component / total and then multiply by 100

•There are a couple of procedures used to calculate percent compositions

o To get the information found on food labels the chemists had to know what fraction of the whole was each component

•Component / total and then multiply by 100

•There are a couple of procedures used to calculate percent compositions


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What percentage of Hydrogen and Oxygen

is in Water (H2O)?

What percentage of Hydrogen and Oxygen

is in Water (H2O)?

Calculating PC given formula Calculating PC given formula

Assume you have 1 mole of water, and calculate its molar mass

Assume you have 1 mole of water, and calculate its molar mass

(2•1.008g) +(2•1.008g) +(1•15.994g) =(1•15.994g) =18.01g18.01g

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oThere are 2 mols of H atoms for every 1 mol of Water molecules

oThere are 2 mols of H atoms for every 1 mol of Water molecules

H: (2•1.008g)= H: (2•1.008g)= 2.016g H 2.016g H

oPercent of H in Water?oPercent of H in Water?

2.016g H2.016g H

18.01 g H2O18.01 g H2OX 100%=X 100%=11.2%11.2%


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oThere is 1 mol of O atoms for every 1 mol of Water molecules

oThere is 1 mol of O atoms for every 1 mol of Water molecules


O: (1•15.994g)= O: (1•15.994g)= 15.994g O 15.994g O

oPercent of O in Water?oPercent of O in Water?15.994 O 15.994 O

18.01 g H2O18.01 g H2OX 100%=X 100%=88.8%88.8%

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o Another method of calculating the percent composition is by experimental analysis.•the overall mass of the sample is measured. Then the sample is separated into its component elements

•The equation is the SAME as before!

o Another method of calculating the percent composition is by experimental analysis.•the overall mass of the sample is measured. Then the sample is separated into its component elements

•The equation is the SAME as before!


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o The masses of the component elements are then determined and the percent composition is calculated as before

•by dividing the mass of each element by the total mass of the sample then multiplying by 100

o The masses of the component elements are then determined and the percent composition is calculated as before

•by dividing the mass of each element by the total mass of the sample then multiplying by 100


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Find the percent composition of a compound

that contains 1.94g of carbon, 0.48g of Hydrogen,

and 2.58g of Sulfur in a 5.0g sample of the

compound.

Find the percent composition of a compound

that contains 1.94g of carbon, 0.48g of Hydrogen,

and 2.58g of Sulfur in a 5.0g sample of the

compound.

Calculating PC given sample Calculating PC given sample

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C: 1.94g/5.0g X 100% = 38.8%C: 1.94g/5.0g X 100% = 38.8%H: 0.48g/5.0g X 100% = 9.6%H: 0.48g/5.0g X 100% = 9.6%

S: 2.58g/5.0g X 100% = 51.6%S: 2.58g/5.0g X 100% = 51.6%

o Calculate the percentage for each element much like you would calculate the percentage for anything.

o Calculate the percentage for each element much like you would calculate the percentage for anything.

Calculating PC given sample Calculating PC given sample

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o Once the percent compositions are determined then they can be used to calculate a simple chem formula for the cmpnd

•key is to convert the percents by mass into amounts in moles

•Then, compare the moles using ratios to determine subscripts

o Once the percent compositions are determined then they can be used to calculate a simple chem formula for the cmpnd

•key is to convert the percents by mass into amounts in moles

•Then, compare the moles using ratios to determine subscripts

Empirical FormulasEmpirical Formulas

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oSince we have been given per-cents rather than masses we need to make an assumption.

•Assume we have a total sample that weighs 100 g.

oSince we have been given per-cents rather than masses we need to make an assumption.

•Assume we have a total sample that weighs 100 g.

What is the empirical formula of a compound that is 80%C and 20%H by mass

What is the empirical formula of a compound that is 80%C and 20%H by mass

Calculating Empirical Formulas


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o This allows us to say that if we had a 100 grams of sample,•80 g is Carbon•20 g is Hydrogen

o Now that we have a set of masses we need to convert them to moles

•Divide by the molar masses from the Periodic Table

o This allows us to say that if we had a 100 grams of sample,•80 g is Carbon•20 g is Hydrogen

o Now that we have a set of masses we need to convert them to moles

•Divide by the molar masses from the Periodic Table

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80g C80g C1 mole C1 mole C

12 g C12 g C==6.7mol C6.7mol C



20g H20g H1 mole H1 mole H

1 g H1 g H==20 mol H20 mol H

•Now calculate the simplest ratio of each by dividing both values by the smallest value

•Now calculate the simplest ratio of each by dividing both values by the smallest value

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Divide each mole value by the smaller of the two values:

Divide each mole value by the smaller of the two values:

C: 6.7/6.7=1C: 6.7/6.7=1H: 20/6.7 = 2.98 3H: 20/6.7 = 2.98 3Ratio is 1 C’s for every 3

H’s; so the formula is =Ratio is 1 C’s for every 3 H’s; so the formula is =CH3CH3



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Determine the empirical formula of a compound

containing 25.9g of N and 74.1g of O.

Determine the empirical formula of a compound

containing 25.9g of N and 74.1g of O.



Notice we have masses this time not percents, we can convert masses

directly to moles

Notice we have masses this time not percents, we can convert masses

directly to moles

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25.9g N25.9g N1 mol N1 mol N

14 g N14 g N==1.85 mol N1.85 mol N



74.1g O74.1g O1 mol O1 mol O

16 g O16 g O==4.63 mol O4.63 mol O

1.85 mol1.85 mol

1.85 mol1.85 mol

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Is the final answer N1O2.5? Of course not!

Is the final answer N1O2.5? Of course not!

We need a whole number ratio…

We need a whole number ratio…

Each part of the ratio is multiplied by a number that

converts the fraction to a whole number

Each part of the ratio is multiplied by a number that

converts the fraction to a whole number

N2(1)O2(2.5)= N2O5N2(1)O2(2.5)= N2O5

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oThe empirical formula indicates the simplest ratio of the atoms in the compnd•However, it does not tell you the actual numbers of atoms in each molecule of the compnd

•For instance, glucose has the molecular formula of C6H12O6

•Empirical form would be CH2O

oThe empirical formula indicates the simplest ratio of the atoms in the compnd•However, it does not tell you the actual numbers of atoms in each molecule of the compnd

•For instance, glucose has the molecular formula of C6H12O6

•Empirical form would be CH2O

Molecular FormulasMolecular Formulas

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oThe empirical formula of CH2O, could be several compnds.•C2H4O2 or C3H6O3 or C100H200O100

oIt’s more important to know the exact numbers of atoms involvedThe numbers of atoms define the properties of the compnd

oThe empirical formula of CH2O, could be several compnds.•C2H4O2 or C3H6O3 or C100H200O100

oIt’s more important to know the exact numbers of atoms involvedThe numbers of atoms define the properties of the compnd


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oThe molecular formula is always a whole-number multiple of the emp. formula

oIn order to calculate the molecular formula you must have 2 pieces of information•Empirical formula•Molar mass of the unknown compound (must be given)

oThe molecular formula is always a whole-number multiple of the emp. formula

oIn order to calculate the molecular formula you must have 2 pieces of information•Empirical formula•Molar mass of the unknown compound (must be given)


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Find the molecular formula of a compound that contains

56.36 g of O and 54.6 g of P. If the molar mass of the compound is 189.5 g/mol.

Find the molecular formula of a compound that contains

56.36 g of O and 54.6 g of P. If the molar mass of the compound is 189.5 g/mol.

1) Find the Empirical Formula2) Find the MM of the Emp.

Form.3) Find the ratio of the 2 molar

masses (Mol MM/Emp MM)

1) Find the Empirical Formula2) Find the MM of the Emp.

Form.3) Find the ratio of the 2 molar

masses (Mol MM/Emp MM)

Calculating Molecular Formulas


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56.36g O56.36g O

54.6g P54.6g P

16 g O16 g O

1 mol O1 mol O= 3.5 mol O= 3.5 mol O

31g P31g P

1 mol P1 mol P= 1.8 mol P= 1.8 mol P

1)Find the Empirical Formula1)Find the Empirical Formula

1.8 mol1.8 mol

1.8 mol1.8 mol

Empirical formula: P1O2Empirical formula: P1O2

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MM of PO2: (1•31g P)MM of PO2: (1•31g P)

189.5 g/mol189.5 g/mol

2)Find the MM of the Emp Form.

2)Find the MM of the Emp Form.

+ (2•16g O)+ (2•16g O)= 63g/mol

3)Find the ratio of the 2 molar masses (mol MM/emp MM)

3)Find the ratio of the 2 molar masses (mol MM/emp MM)

63 g/mol63 g/mol= 3.00= 3.00

GIVEN GIVEN GIVEN GIVEN

CALCULATED CALCULATED CALCULATED CALCULATED

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oSo the Molecular formula is 3 times heavier than the Empirical formula•Therefore, the molecular formula has 3 times more atoms than the emp. formula

oSo the Molecular formula is 3 times heavier than the Empirical formula•Therefore, the molecular formula has 3 times more atoms than the emp. formula



P3(1)O2(3)= P3O6P3(1)O2(3)= P3O6

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1 PERCENT COMPOSITION, EMPIRICAL & MOLECULAR FORMULAS