Chemical Composition-Chapter 6

Chemical Composition

• Counting by Weighing– Jelly beans are not identical in mass – Suppose you measured the mass of 10 jelly

beans– 5.1g, 5.2g, 5.0g, 4.8g, 4.9g, 5.0g, 5.0g, 5.1g,

4.9g, 5.0g


• We can use these masses to calculate the average mass

• Average Mass = Total Mass of Beans

Number of Beans

• Average Mass = 50.0 = 5.0g

10

• Therefore to measure 100 jelly beans you would weigh 500g of jelly beans


• If a mint has an average mass of 15g how do the jelly beans compare to the mints?

• Sample contains 1 component – Mints 15g– Jelly Beans 5g

• Sample contains 10 components– Mints 150g– Jelly Beans 50g

• Ratio is always 3:1, mints 3 times as massive


• A marble has an average mass of 8g, what mass of marbles would I have to measure to count out 25 marbles?


• Atomic Masses– Counting atoms by weighing

• C(s) + O2(g) CO2(g)

• The above equation tells us that one atom of carbon reaction with one molecule of oxygen to form 1 molecule of carbon dioxide


• To determine how many oxygen molecules are required we must know how many carbon atoms are present

• We must learn to count atoms by weighing

• Atoms are much too small to use units of measurement such as grams or kilograms

• The mass of a single carbon atom is 1.99 x 10-23g


• To avoid using very small numbers when describing an atom scientists have defined a much smaller mass unit called the atomic mass unit (amu)

• 1 amu = 1.66 x 10-24g

• In order to count carbon atoms by weighing we must know the average atomic mass


• The average atomic mass is the average of the masses of the different isotopes of each element

• The average atomic mass of carbon is 12.01amu

• This means that a sample of carbon can be treated as though it is made up of identical atoms each with a mass of 12.01


• What mass of natural carbon must was take to have 1000 atoms present?

• Mass of 1000 carbon atoms =

(1000 atoms)(12.01 amu)

atom

• Mass of 1000 carbon atoms=

1.204 x 104 amu


• 1 carbon atom = 12.01 amu

• Sample of carbon weighs 3.00 x 1020amu

• 3.00 x 1020 amu x 1 carbon atom =

12.01 amu

• 2.50 x 1019 carbon atoms


• Calculate the mass of a sample of aluminum that contains 75 atoms

• Calculate the number of sodium atoms in a sample that has a mass of 1172.49 amu


• Calculate the mass of a sample that contains 23 nitrogen atoms

• Calculate the number of oxygen atoms in a sample that has a mass of 288 amu


• Atomic Mass Units (amu) are very small units

• We must learn to count atoms in samples by using masses measured in grams

• We use the mole to describe the number of atoms in a specific mass (grams) of a particular element


• Mole: the number equal to the number of carbon atoms in a 12.01g sample of carbon

• Avagadro’s Number: 6.022 x 1023, a mole of something contains 6.022 x 1023 units of that substance


• How do we use moles in calculations?– Recall: 1 mole is defined as the number of

atoms in 12.01g of carbon– Therefore there are 6.022 x 1023 atoms in

12.01g of Carbon– The average atomic mass of carbon is

12.01amu– A sample of any element that weighs a

number of grams equal to the atomic mass has 6.022 x 1023 atoms present


Element Number of Atoms

Mass of Sample (g)

Aluminum 6.022 x 1023 26.98

Gold 6.022 x 1023 196.97

Iron 6.022 x 1023 55.85

Sulfur 6.022 x 1023 32.07

Boron 6.022 x 1023 10.81

Xenon 6.022 x 1023 131.3


• Sample A contains a mole of Hydrogen atoms, sample A has a mass of 1.008g. Sample B has an unknown amount of atoms of Hydrogen. How can we determine the number of atoms in sample B?– Assume the mass of Sample B is 0.500g


• 0.500g H x 1 mole H = 0.496 mole H 1.008g H

• We know that 1 mole = 6.022 x 1023 atoms

• 0.496 mole H x 6.022 x 1023 H atoms = 1 mole H

• 2.99x1023 H atoms in Sample B


• Aluminum metal is often used in structures such as bicycle frames. Compute the number of moles and atoms in a 10g sample of aluminum.


• A silicon chip used in an integrated circuit of a microcomputer has a mass of 5.68g. How many silicon atoms are present in this chip? The average atomic mass of silicon is 28.09amu.


• Chromium (Cr) is a metal that is added to steel to improve its resistance to corrosion (for example to make stainless steel). Calculate both the moles in a sample of chromium containing 5.00 x 1020 atoms and the mass of the sample.


• A chemical compound is a collection of atoms

• CH4

– Consists of 1 C atom and 4 H atoms

• How can we calculate the mass of 6.022 x 1023 molecules?


• Since the compound CH4 contains 1 C atom and 4 H atoms

• 1 mole of CH4 contains 1 mole of C atoms and 4 moles of H atoms


• 10 CH4 atoms would contain 10 Carbon atoms and 40 Hydrogen atoms

• 1 Mole of CH4 would contain 6.022 x 1023 carbon atoms and 4(6.022 x 1023) hydrogen atoms

• Mass CH4 = 12.01g + 4(1.008g)

• Mass CH4 = 16.04g


• Molar Mass: the mass of 1 mole of a substance, obtained by summing the masses of the component atoms


• Calculate the molar mass of sulfur dioxide, a gas produced when sulfur containing fuels are burned. – Formula

• SO2

– How many S atoms in 1 molecule?• 1

– How many O atoms in 1 molecule?• 2

– 1(32.07g) + 2(15.999g)= 64.07g


• Calcium carbonate, CaCO3 (also called calcite), is the principle material found in limestone, marble, chalk, pearls, and the shells of marine animals– Calculate the molar mass of calcium

carbonate– A certain sample of calcium carbonate

contains 4.86mole. What is the mass in grams of this sample?


• Juglone, a dye known for centuries, is produced from the husks of black walnuts. It is also a natural herbicide (weed killer) that kills off competitive plants around the black walnut tree but does not affect grass and other noncompetitive plants. The formula of Juglone is C10H6O3.

– Calculate the molar mass of Juglone– A sample of 1.56g of pure juglone was extracted from

the black walnut husks. How many moles of juglone does this sample represent?


• Isopentyl acetate, C7H14O2, the compound responsible for the scent of bananas, can be produced commercially. This compound is also released when bees sting in the amount of 1 x 10-6g. How many moles and how many molecules of the isopentyl acetate are released in a typical bee sting?


• Mass Percent: defines the mass of a component of a substance relative to the total mass of the substance

• Mass Percent = Mass of Part X 100%

Mass of Whole


Given the compound C2H5OH we can calculate the mass percent of each component. First, we must calculate the molar mass.

• Molar mass=2(12.01) + 6(1.008) + 15.999 – 46.07g


• Mass of C = 2(12.01g) = 24.02g

• Mass percent of C– 24.02g x 100% = 52.14%

46.07g

• Mass of H = 6(1.008g) = 6.048g

• Mass percent of H – 6.048g x 100% = 13.13%

46.07g


• Carvone is a substance that occurs in two forms, both of which have the same molecular formula (C10H14O) and molar mass. One type of carvone gives caraway seeds their characteristic smell; the other is responsible for the smell of peppermint oil. Compute the mass percent of each element in carvone.


• Empirical Formula: the simplest formula of a compound, the smallest whole number ratio of a compound

• Molecular Formula: the actual formula of a compound, the one that gives the actual composition of the molecules present


• Determining Empirical Formulas:

– C6H6

• CH

– C12H4Cl4O2

• C6H2Cl2O

– C6H16N2

• C3H8N


• Steps for Determining the Empirical Formula of a Compound:– Obtain the Mass of each element– Determine the number of moles of each – Divide the number of moles of each element

by the number of moles of the smallest – Multiply all the numbers you obtained by the

smallest integer that will convert all the numbers to whole numbers


• An oxide of aluminum is formed by the reaction of 4.151g of aluminum with 3.692g of oxygen. Calculate the empirical formula of this compound.


• When a 0.3546g sample of vanadium metal is heated in air, it reacts with oxygen to achieve a final mass of 0.6330g. Calculate the empirical formula of this vanadium oxide.


• A sample of lead arsenate, an insecticide used against the potato beetle, contains 1.3813g of lead, 0.00672g of hydrogen, 0.4995g of arsenic, and 0.4267g oxygen. Calculate the empirical formula for lead arsenate.


• Cisplatin, the common name for a platinum compound that is used to treat cancerous tumors, has the composition (mass percent) 65.02% platinum, 9.34% nitrogen, 2.02% hydrogen, and 23.63% chlorine. Calculate the empirical formula for cisplatin.


• A white powder is analyzed and found to have an empirical formula of P2O5. The compound has a molar mass of 283.88. What is the compound’s molecular formula?

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Chemical Composition-Chapter 6