Molecules, Compounds, & Chemical Equations. Overview Empirical vs. Molecular Formulas Mass Percent...

Molecules, Compounds, & Chemical Equations

Overview

• Empirical vs. Molecular Formulas• Mass Percent• Determining Chemical Formulas

Empirical vs. Molecular Formulas

• Empirical Formula represents the simplest whole number ratio of atoms in a formula.

• Molecular Formula represents the true formula for a substance and is usually a multiple of the empirical formula.

Mass Percent

• Empirical Formulas are determined from mass percent data.

• To find the mass percent of an element use the following equation.

Mass % of X = (mass X / total mass of compound) x 100%

• The mass percentages for all elements in a compound must sum to 100.00%.

Mass Percent

• Find the mass % oxygen in sodium nitrate (NaNO3). Molar mass of NaNO3 is 85.07 g/mole.

Mass Percent• Given:

sodium nitrate NaNO3

Molar Mass = 85.07 g/mole

• Unknown: ? Mass % oxygen

• Relevant Information:

1 mole Na, 1 mole N, and 3 moles O are found in NaNO3

Mass Percent X = (mass X / total mass of compound) x 100%

Mass Percent

• Solution:

mass O = (3 mol x 16.00 g/mol) = 48.00 g

Total mass (NaNO3) = (1 mol x 23.00 g/mol) + (1 mol x 14.07 g/mol) + (3 mol x 16.00 g/mol) = 85.07 g

Mass % O = (48.00 g / 85.07 g) x 100% = 56.42% O

• Check: Answer is reasonable.

Mass Percent

• For additional practice, find the % Na and % N in sodium

nitrate.

Answers: 27.04% Na16.54% N

Determining Chemical Formulas

• To find an empirical formula, one would use the following method (Works Every Time).


• Example Problem

– 71.65% Cl Mol. Wt. = 98.96 g/mol– 24.27% C– 4.07% H

– Find the empirical & molecular formulas for this compound.


• The Traditional Approach

– Step 1: Assume 100.00 g sample

• We pick this sample size so that our percentage data will translate directly into mass in grams.

71.65% Cl 71.65 g Cl24.27% C 24.27 g C 4.07% H 4.07 g H



– Step 2: Convert mass values to moles (Axiom #3)

71.65 g Cl x ( 1 mol Cl / 35.45 g Cl) = 2.021 mol Cl

24.27 g C x ( 1 mol C / 12.01 g C) = 2.021 mol C

4.07 g H x ( 1 mol H / 1.008 g H) = 4.04 mol H


• Traditional Approach (Works Every Time)

– Step 3: Divide number of moles obtained from step 2 by smallest number of moles present. This gives the mole ratios for each component (look at this value as being the same thing as the number of atoms present).



– Step 3: Convert moles to mole ratios (# atoms)

2.021 mol Cl / 2.021 mol = 1.000 Cl atoms

2.021 mol C / 2.021 mol = 1.000 C atoms

4.04 mol H / 2.021 mol = 2.00 H atoms


• Traditional Approach (Works Every Time)

– Step 4: Examine values obtained from step 3. The mole ratios should be whole numbers. If you have whole numbers, then you have the subscripts for the empirical formula directly. If not, proceed to step 5.

• Note: Values like 2.98, 4.03, etc. translate into 3.00 and 4.00, respectively. Do not round numbers like 2.5, 3.67, 1.33, etc. to whole numbers.



– Step 4: Values are whole numbers, so we can now write the empirical formula

1.000 Cl atoms

1.000 C atoms Empirical Formula = CH2Cl

2.00 H atoms


• Traditional Method (Works Every Time)

– Step 5: If mole ratios are not whole numbers, then one must introduce a factor that will convert the mole ratios into whole numbers.

– Once you have found the whole number ratios, then the subscripts for the empirical formula are then known. Simply write out the empirical formula.


• To find a molecular formula, one would use the same procedure as the empirical formula with the following additional steps.

– Step 6: Once you have empirical formula, calculate the empirical molar mass (EMM).



– Step 6: Calculate Empirical Molar Mass (EMM)

Empirical Formula = CH2Cl

EMM = (1 x 12.01 g/mol) + (2 x 1.01 g/mol) + (1 x 35.45 g/mol) = 49.48 g/mol


• To find a molecular formula

– Step 7: Using the molecular weight (this must be given in problem), find the factor (x) that must be multiplied through to get molecular formula.

EMM (x) = Mol. Wt.

(Empirical Formula) (x) => Molecular Formula



– Step 7: Determine Factor (x)

EMM (x) = Mol. Wt.

49.48 g/mol (x) = 98.96 g/mol (x) = 2.000 = 2

Molecular Formula = (CH2CL) (2) => C2H4Cl2

Molecules, Compounds, & Chemical Equations. Overview Empirical vs. Molecular Formulas Mass Percent...

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