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# Copy of Percent Composition, Emprical & Molecular Formulas !

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Tweet## David Maxin

on 19 December 2015#### Transcript of Copy of Percent Composition, Emprical & Molecular Formulas !

The Composition of Unknown Compounds 6.6 Percent Composition Law of Definite Proportions A chemical compound always contains exactly the same proportion of elements by mass. FOR EXAMPLE Carbon would always combine in the same proportions with oxygen to produce carbon dioxide (CO2) What is it? Percent composition is the percent of the total mass each element has in a compound. The mass percent of each element in a compound can be found using the formula: SAMPLE PROBLEM ONE Calculate the percent composition of a compound

that is made of 29.0 g of Ag with 4.30 g of S.

Total mass = 29.0 + 4.30 = 33.3 g Remember!! The sum of % composition for each element should equal 100% SAMPLE PROBLEM TWO Determine the percent composition of calcium in Ca3(PO4)2.

First calculate the mass of 1 mole of Ca3(PO4)2.

Mass of Ca = 3 mol x 40.1 g/mol = 120.3 g

Mass of P = 2 mol x 31.0 g/mol = 62.0 g

Mass of O = 8 mol x 16.0 g/mol = 128.0g

Total mass = 310.3 g Chemical Formulas of Compounds

Empirical and Molecular Formula give the relative numbers of atoms or moles of each element in a formula unit.

Example: CO2 has two atoms of oxygen for every one atom of carbon. TWO TYPES OF FORMULAE 1. Empirical Formula The formula of a compound that represents a mole ratio in the smallest whole number ratio of the atoms present. For example: The Empirical Formula of C2H4 is CH2 How to obtain an Empirical Formula? Follow these FOUR steps!

1.Determine the mass in grams of each element present, if necessary.

2. Calculate the number of moles of each element.

3. Divide each by the smallest number of moles to obtain the simplest whole number ratio.

4. If whole numbers are not obtained* in step 3, multiply through by the smallest number that will give all whole numbers A chart is helpful !! A compound of carbon, chlorine and fluorine was analyzed and found to contain 16.3% carbon, 32.1% chlorine, 51.6% fluorine by MASS. Determine the simple formula of the compound. Therefore, empirical formula is C3Cl2F6 2. Molecular Formula The actual ratio of the number of atoms of each type bonded to form a molecule For Example: The Molecular Formula of CH2 is C2H4 How do we obtain a Molecular Formula? To determine molecular formula you must know

Empirical formula

Molar mass of compound SAMPLE PROBLEM ONE A student has determined that the empirical formula for a compound of sulphur and chlorine is SCl. The molar mass of this compound is 135.0 g/mol. Determine its molecular formula.

Find M of empirical formula. MSCl = 32.1 + 35.5 = 67.6 g/mol

Divide M of compound by M of empirical formula to get the # of units.

∴Simple formula = SCl x2 S2Cl2 = molecular formula From Empirical Molecular Determine the molecular formula of a compound containing 85.7% carbon and 14.3% hydrogen by mass. The molar mass of the compound is 84.0 g/mol

M(CH2) = 2(1.0) + 12.0 = 14.0 g/mol

M compound = 84.0 g/mol

SO Molecular formula is 6 x (CH2) = C6H12

Full transcriptthat is made of 29.0 g of Ag with 4.30 g of S.

Total mass = 29.0 + 4.30 = 33.3 g Remember!! The sum of % composition for each element should equal 100% SAMPLE PROBLEM TWO Determine the percent composition of calcium in Ca3(PO4)2.

First calculate the mass of 1 mole of Ca3(PO4)2.

Mass of Ca = 3 mol x 40.1 g/mol = 120.3 g

Mass of P = 2 mol x 31.0 g/mol = 62.0 g

Mass of O = 8 mol x 16.0 g/mol = 128.0g

Total mass = 310.3 g Chemical Formulas of Compounds

Empirical and Molecular Formula give the relative numbers of atoms or moles of each element in a formula unit.

Example: CO2 has two atoms of oxygen for every one atom of carbon. TWO TYPES OF FORMULAE 1. Empirical Formula The formula of a compound that represents a mole ratio in the smallest whole number ratio of the atoms present. For example: The Empirical Formula of C2H4 is CH2 How to obtain an Empirical Formula? Follow these FOUR steps!

1.Determine the mass in grams of each element present, if necessary.

2. Calculate the number of moles of each element.

3. Divide each by the smallest number of moles to obtain the simplest whole number ratio.

4. If whole numbers are not obtained* in step 3, multiply through by the smallest number that will give all whole numbers A chart is helpful !! A compound of carbon, chlorine and fluorine was analyzed and found to contain 16.3% carbon, 32.1% chlorine, 51.6% fluorine by MASS. Determine the simple formula of the compound. Therefore, empirical formula is C3Cl2F6 2. Molecular Formula The actual ratio of the number of atoms of each type bonded to form a molecule For Example: The Molecular Formula of CH2 is C2H4 How do we obtain a Molecular Formula? To determine molecular formula you must know

Empirical formula

Molar mass of compound SAMPLE PROBLEM ONE A student has determined that the empirical formula for a compound of sulphur and chlorine is SCl. The molar mass of this compound is 135.0 g/mol. Determine its molecular formula.

Find M of empirical formula. MSCl = 32.1 + 35.5 = 67.6 g/mol

Divide M of compound by M of empirical formula to get the # of units.

∴Simple formula = SCl x2 S2Cl2 = molecular formula From Empirical Molecular Determine the molecular formula of a compound containing 85.7% carbon and 14.3% hydrogen by mass. The molar mass of the compound is 84.0 g/mol

M(CH2) = 2(1.0) + 12.0 = 14.0 g/mol

M compound = 84.0 g/mol

SO Molecular formula is 6 x (CH2) = C6H12