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Empirical Formulas & Percent Composition. CO 2 Carbon dioxide. BCl 3 boron trichloride. Molar Mass. How many moles of alcohol are there in a “standard” can of beer if there are 21.3 g of C 2 H 6 O?. Molar mass of C 2 H 6 O = 46.08 g/mol Calc. moles of alcohol.
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Empirical Formulas & Percent Composition CO2 Carbon dioxide BCl3 boron trichloride
How many moles of alcohol are there in a “standard” can of beer if there are 21.3 g of C2H6O? • Molar mass of C2H6O = 46.08 g/mol • Calc. moles of alcohol
How many moleculesof alcohol are there in a “standard” can of beer if there are 21.3 g of C2H6O? = 2.78 x 1023 molecules We know there are 0.462 mol of C2H6O.
How many atoms of Care there in a “standard” can of beer if there are 21.3 g of C2H6O? = 5.57 x 1023 C atoms There are 2.78 x 1023 molecules. Each molecule contains 2 C atoms. Therefore, the number of C atoms is
Empirical & Molecular Formulas A pure compound always consists of the same elements combined in the same proportions by weight. Therefore, we can express molecular composition as PERCENT BY WEIGHT Ethanol, C2H6O 52.13% C 13.15% H 34.72% O
Structure of NO2 Percent Composition Consider some of the family of nitrogen-oxygen compounds: NO2, nitrogen dioxide What is the weight percent of N and of O?
Percent Composition 46 g/ mol Consider NO2, Molar mass = Percent of N and of O? What are percentages of N and O in NO?
Determining Formulas In chemical analysis we determine the % by weight of each element in a given amount of pure compound and derive the EMPIRICALor SIMPLESTformula. PROBLEM: What is its empirical formula of a compound of B and H if 81.10% is boron?
Determination of empirical formulas:mass of each %B + %H = 100% 100% - 81.10% B =18.90% H. If there were 100.0 g of the compound: there are 81.10 g of B and 18.90 g of H. Calculate the number of MOLES of each.
Determination of empirical formulas:moles of each Calculate the number of moles of each element assuming 100.0 g total.
Determination of empirical formulas:moles ratio Take the ratio of moles of B and H. atoms combine in small whole number ratios Always divide by the smaller number. 18.75 mol H and 7.502 mol B 7.502 mol 7.502 mol 2.5 mol H to 1.0 mol B
EMPIRICAL FORMULA = B2H5 But we need a whole number ratio. 2.5 mol H or 5 mol H 1.0 mol B 2 mol B
If the empirical formula is B2H5, what is its molecular formula? We need to do an EXPERIMENT to find the MOLAR MASS. MS gives 53.3 g/mol Compare empirical mass = 26.66 g/unit Find the ratio of these masses. Molecular formula = B4H10
Determine the formula of a compound of Sn and I using the following data. • Reaction of Sn and I2 is done using excess Sn. • Mass of Sn in the beginning = 1.056 g • Mass of Sn remaining = 0.601 g • Mass of iodine (I2) used = 1.947 g
Tin and Iodine Compound Find the mass of Sn that combined with 1.947 g I2. Mass of Sn initially = 1.056 g Mass of Sn recovered = 0.601 g Mass of Sn used = 0.455 g Find moles of Sn used:
Tin and Iodine Compound Now find the number of moles of I2that combined with 3.83 x 10-3 mol Sn. Mass of I2 used was 1.947 g. How many mol of iodine atoms? = 1.534 x 10-2 mol I atoms
Tin and Iodine Compound Now find the ratio of number of moles of moles of I and Sn that combined. Empirical formula is SnI4
Practice Formula Determination Homework: Worksheets pgs 17-20 Pg 13-14 good review for mole calculations Prelab questions