empirical formula & molecular formula. empirical formulas best way to identify an unknown...

Empirical Formula & Molecular

Formula

Empirical Formulas

•Best way to identify an unknown compound is to determine its chemical formula.

•Empirical Formula: the smallest whole number ratio of a compound. (Ionic compounds are already in its empirical formulas)

Empirical Vs. Molecular Formulas

•Molecular Formula

• a formula that shows the element symbols & exact number of each type of atom in a molecular compound

• Empirical Formula

• a formula that shows the simplest whole-number ratio of elements in compounds

CH4 Vs. Vs. C18H72C6H24

Lowest ratioIn empirical

& molecular formula

can be reduced to lowest ratio6 can go into 6 & 24--> CH4

same empirical formula. But not the

same molecular formula

can be reduced to lowest ratio18 can go into

18 & 72--> CH4

•If you know the chemical formula you can find percent composition.

CH4 C6H24

M= 12.01g/mol + (4 x 1.01g/mol) = 16.00 g/mol

MC= 12.01g/mol

MH= 1.01g/mol

(12.00g/mol )x 1mol

16.00g

x100% = 75%

(4.01g/mol)x 1mol

16.00g

x 100% = 25%

MC6H24= 96 g/mol

MC= 72 g/mol

(72 g/mol ) 1molx96.00g

x100%= 75%

- same empirical formula- should be ratios of each other-helps determine percent composition

CH4 C6H24&

Determining Empirical Formula• Find the empirical formula of a compound

with percentage composition 35.4% Na and the remainder nitrogen

Step 1: Assume your conveniently working with a mass of 100.0 g.

100.0g - 35.4g = 64.6 g

Na= 35.4 g

N= 64.6 gStep 2: Calculate the amount of each element.

nNa = (35.4 g) (1 mol)22.99g

nN = (64.5 g) (1 mol)14.01g

= 1.540 mol = 4.6110 mol



Step 3: To determine the simplest ratio of the elements in the compound, divide the amount of each element by the smallest amount

nNa

= 1

nN

1.540 mol4.6110 mol

nNa= 1.540 mol

nNa= 1.540 mol

= 2.98



Step 2: Calculate the amount of each element.

nNa

= 1nN 1.540 mol

4.6110 molnNa

= 1.540 mol nNa

= 1.540 mol

= 2.98The ratio for Na & N is1:2.98. Since we cannot have a fraction of an element in a compound, the value of N is rounded off to the nearest whole number. As a result, the simplest whole-number ratio is 1:3........NaN3

•If a number is within 0.05 of a whole number, you can round it up or down to the nearest whole number.

•But what do you do if one of the values is not close to a whole number? You multiply all the subscripts by the fraction denominator gives whole numbers

•C1 x 3 H1.333 x 3 = C3H4

Practice• Determine the empirical formula of a compound listed that

contains 52.2% carbon, 6.15% hydrogen and 41.7% oxygen

Step 1: Assume your conveniently working with a mass of 100.0 g.

nc =(52.2 g) (1 mol)12.01g

= 4.3464 mol

nH =(6.15g) (1 mol)1.01g

= 6.0891mol

n0 =(41.7g) (1 mol)16.00g

= 2.6063mol

Practice• Determine the empirical formula of a compound listed that

contains 52.2% carbon, 6.15% hydrogen and 41.7% oxygen

Step 2: Divide the amount of each element by the smallest amount.

nc =

= 1

nH =

= 1.67

n0 =2.6063moln0

n0

n0

2.6063mol

2.6063mol

2.6063mol6.0891mol

4.3464 mol

= 2.29

Step 3: These calculations give an empirical formula of C1.67H2.29O1

Multiplying each of the subscripts by 3 give C5H7O3

Practice• Write the empirical formula of compounds with the

following percentage

• 20.2% Al; 79.8% Cl

Practice• Write the empirical formula of compounds with the

following percentage

• 50.85% carbon; 8.47% hydrogen; and 40.68% oxygen

Molecular Formula•gives the exact number of atoms of

each element present

•“chemical formula” = molecular formula

•A compound’s molar mass is always a whole number multiple of the molar mass of the empirical formula. You ca find this multiple x, by dividing the molar mass of the compound by the molar mass of the empirical formula

Molecular Formula

x = molar mass of compoundmolar mass of empirical

MCH2 = 1( 12.01 g/mol) + 1(1.01g/mol)

Find a Molecular Formula Given its Empirical formula

Step 1: Find the empirical molar mass by adding the molar mass of each element

Compound with empirical formula CH2 & the molar mass of 84.18 g/mol

= 14.03 g/mol

x = 84.18 g/mol14.03 g/mol


Step 2: Solve for x,

Compound with empirical formula CH2 & the molar mass of 84.18g/mol

x = 6

CH2= C6H12


Step 3: therefore, the molar mass of the compound is 6 times the molar mass of the

empirical formula

Compound with empirical formula CH2 & the molar mass of 84.18g/mol

molecular formula

Practice• Determine the molecular formulas

• KSO4; 270.32 g/mol

Using Percentage Composition & Molar

Mass Data•Determine the molecular formula of vitamin C

(ascorbic acid). This compound contains 40.5% Carbon, 4.6% hydrogen, 54.5% oxygen. Its molar mass is 176.14g/mol

empirical formula & molecular formula. empirical formulas best way to identify an unknown...

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