How many moles are in 5.5 x 1025 atoms of S? 2. What is the mass in g of 1.505 x 1024 atoms of Cl? 3. How many moles of Ne are in 25.0 g of Ne? 4.
Chemical Composition: Molar Mass and Formulas Chem 101 Lectures 12 and 13
What can we tell from a balanced equation? C + O2
CH4 + 2O2 CH4(g) + 2O2(g)
CO2
CO2 + 2H2O CO2(g) + 2H2O(l)
What else can we figure out from a balanced equation? • If you know how many C atoms at start – how many O2 molecules you will need – how many CO2 molecules you can make
• Count the atoms • Weigh the carbon and calculate the number of atoms.
• Atomic masses allow us to convert weights into numbers of atoms.
Moles • The mass of 1.0 mole of an element is equal to the atomic mass in grams. • One mole element = 6.022 x 1023 atoms. This number is called Avogadro’s number. • Example: 1 mole of C atoms weighs 12.0 g and has 6.02 x 1023 atoms.
A 1-mol sample of graphite (a form of carbon) weighs 12.01 g.
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Equalities • For any element: • 1 mole = 6.02 x 1023 atoms = Atomic mass • 1 mole of calcium = – 6.02 x 1023 atoms of calcium – 40.078 grams of calcium
• 1 mole of oxygen (O) atoms = – 6.02 x 1023 atoms of oxygen – 15.999 grams of oxygen
• 1 mole of oxygen gas (O2) =
– 12.04 x 1023 atoms of oxygen – 31.998 grams of oxygen
All of these examples of pure elements contain the same number (a mole) of atoms: 6.02 x 1023 atoms.
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One-mole samples of iron (nails), iodine crystals, liquid mercury, and powdered sulfur.
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How many moles and atoms are in 10.0 g of Al? 1. What are the equalities?
2. Use the atomic mass as a conversion factor for grams-to-moles.
3. Use Avogadro’s number as a conversion factor for moles-to-atoms.
How many moles and grams are in 2.23 x 1023 atoms of Al. 1. What are the equalities? 2. Use Avogadro’s number as a conversion factor for atoms-to-moles.
3. Use atomic mass as a conversion factor for moles-to-grams.
Various numbers of methane molecules and their constituent atoms.
One mole of SO2 contains 1 mole of S and 2 moles of O.
Molar Mass • Molar mass: the mass in grams of one mole of a compound • The relative weights of molecules can be calculated from atomic masses Water = H2O = 2(1.008 g) + 16.00 g = 18.02 g • 1 mole of H2O will weigh 18.02 g, therefore the molar mass of H2O is 18.02 g • 1 mole of H2O will contain 16.00 g of oxygen and 2.02 g of hydrogen
Molar Mass • For any molecule: • 1 mole = 6.02 x 1023 molecules = molar mass • 1 mole of hydrogen gas = 1 mole of H2 – 6.02 x 1023 molecules of H2 – 2.016 grams of H2 – 2 x 6.02 x 1023 atoms of H for a total of 12.04 x 1023 atoms or 1.204 x 1024 atoms
Molar Mass • For any molecule: • 1 mole = 6.02 x 1023 molecules = molar mass • 1 mole of water = 1 mole of H2O – 6.02 x 1023 molecules of H2O – 18.02 grams of H2O – How many atoms?
• 1 mole of hydrogen gas = 1 mole of H2 – 6.02 x 1023 molecules of H2 – 2.016 grams of H2 – How many atoms?
Practice Mole Problems 1.
How many moles are in 5.5 x 1025 atoms of S?
2.
What is the mass in g of 1.505 x 1024 atoms of Cl?
3.
How many moles of Ne are in 25.0 g of Ne?
4.
How many atoms of Na are in 75.0 g of Na?
5. Which contains more atoms: 50.0 g of Al or 50.0 g of Fe?
6. Which contains more Ni: 20.0g 2.85 x 1023 atoms 0.45 mol? 7. How many molecules are in 0.5 moles of CaCl2?
Chemical Composition: Part 2 • Percent composition • Empirical formulas • Molecular formulas
What is Percent Composition? • Percentage by mass of each element in a compound • Can be determined from the formula of the compound or by experimental mass analysis of the compound • The percentages may not always total to 100% due to rounding.
Percentage
part 100% whole
How to determine the percent composition for each element in a formula •
Determine the mass of each element in 1 mole of the compound.
•
Determine the molar mass of the compound by adding the masses of the elements.
•
Divide the mass of each element by the molar mass of the compound and multiply by 100%
What is the percent composition for each element in Ethanol? C2H5OH
Percent composition for elements in C2H5OH • Determine the mass of each element in 1 mole of the compound C2H5OH. • 2 moles C = 2( • 6 moles H = 6( • 1 moles O = 1(
g) = g) = g) =
g g g
• Determine the molar mass of the compound by adding the masses of the elements. • 1 mole C2H5OH =
g
• Divide the mass of each element by the molar mass of the compound and multiply by 100% • C
• H • O
Empirical Formulas • Empirical formula: the simplest, wholenumber ratio of atoms in a molecule – Can be determined from percent composition or by combining masses – Different substances can have the same empirical formula
• Molecular formula: a multiple of the empirical formula
Same Empirical Formula CH2O Formaldehyde
Acetic acid
Dextrose
Benzopyrene, C20H12 • Benzopyrene is found in nature from the eruption of volcanoes and forest fires. It is also produced by burning plants, wood, coal, and operating cars, trucks and other vehicles. • The major indoor sources of benzopyrene in the air are wood-burning fireplaces and stoves, and tobacco smoking. • Benzopyrene can be found in surface water, tap water, rainwater, groundwater, wastewater and sewage sludge. • No known industry production or use.
• Skin and eye irritant, Carcinogenic
Determine the empirical formula of benzopyrene, C20H12 •
Find the greatest common factor (GCF) of the subscripts. factors of 20 = (10 x 2), (5 x 4) factors of 12 = (6 x 2), (4 x 3) GCF = 4
•
Divide each subscript by the GCF to get the empirical formula. C20H12 = (C5H3)4
Empirical Formula = C5H3
Determine the empirical formula of acetic anhydride by percent composition: 47% carbon, 47% oxygen, and 6.0% hydrogen. Convert the percentages to grams by assuming you have 100 g of the compound.
Convert the grams to moles Divide by the smallest number of moles If any of the ratios is not a whole number, multiply all the ratios by a factor to make it a whole number.
Use the ratios as the subscripts in the empirical formula.
Molecular Formulas • The molecular formula is a multiple of the empirical formula. • To determine the molecular formula you need to know the empirical formula and the molar mass of the compound.
Determine the molecular formula of benzopyrene if it has a molar mass of 252 g and an empirical formula of C5H3 • What is the empirical formula: C5H3 • Determine the molar mass of C5H3
• 5 C =_____g, 3 H =_____g, so C5H3 =______g
• Divide the given molar mass of the compound by the molar mass of the empirical formula • Round to the nearest whole number • Multiply the empirical formula by the calculated factor to give the molecular formula (C5H3)__ = C H
Empirical and Molecular Formulas • Different compounds may have the same empirical formula • The molecular formula and structure identifies the different compounds
