The document discusses several chemistry concepts including: - Calculating average mass and ratios using sample masses of jelly beans and mints. - Using atomic mass units (amu) to count atoms by weighing samples and determining the number of atoms in samples of various elements. - Defining the mole as 6.022x10^23 units of a substance, and using moles to calculate the number of atoms in samples. - Calculating molar mass and using molar mass to determine the mass of samples containing a given number of moles.

1. 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

2. Chemical Composition 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

3. Chemical Composition 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

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

5. Chemical Composition Atomic Masses Counting atoms by weighing C (s) + O 2 (g)  CO 2 (g) The above equation tells us that one atom of carbon reaction with one molecule of oxygen to form 1 molecule of carbon dioxide

6. Chemical Composition 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 -23 g

7. Chemical Composition 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 -24 g In order to count carbon atoms by weighing we must know the average atomic mass

8. Chemical Composition 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

9. Chemical Composition 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 10 4 amu

10. Chemical Composition 1 carbon atom = 12.01 amu Sample of carbon weighs 3.00 x 10 20 amu 3.00 x 10 20 amu x 1 carbon atom = 12.01 amu 2.50 x 10 19 carbon atoms

11. Chemical Composition 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

12. Chemical Composition 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

13. Chemical Composition 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

14. Chemical Composition Mole : the number equal to the number of carbon atoms in a 12.01g sample of carbon Avagadro’s Number : 6.022 x 10 23 , a mole of something contains 6.022 x 10 23 units of that substance

15. Chemical Composition 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 10 23 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 10 23 atoms present

16. Chemical Composition 131.3 6.022 x 10 23 Xenon 10.81 6.022 x 10 23 Boron 32.07 6.022 x 10 23 Sulfur 55.85 6.022 x 10 23 Iron 196.97 6.022 x 10 23 Gold 26.98 6.022 x 10 23 Aluminum Mass of Sample (g) Number of Atoms Element

17. Chemical Composition 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

18. Chemical Composition 0.500g H x 1 mole H = 0.496 mole H 1.008g H We know that 1 mole = 6.022 x 10 23 atoms 0.496 mole H x 6.022 x 10 23 H atoms = 1 mole H 2.99x10 23 H atoms in Sample B

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

20. Chemical Composition 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.

21. Chemical Composition 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 10 20 atoms and the mass of the sample.

22. Chemical Composition A chemical compound is a collection of atoms CH 4 Consists of 1 C atom and 4 H atoms How can we calculate the mass of 6.022 x 10 23 molecules?

23. Chemical Composition Since the compound CH 4 contains 1 C atom and 4 H atoms 1 mole of CH 4 contains 1 mole of C atoms and 4 moles of H atoms

24. Chemical Composition 10 CH 4 atoms would contain 10 Carbon atoms and 40 Hydrogen atoms 1 Mole of CH 4 would contain 6.022 x 10 23 carbon atoms and 4(6.022 x 10 23 ) hydrogen atoms Mass CH 4 = 12.01g + 4(1.008g) Mass CH 4 = 16.04g

25. Chemical Composition Molar Mass : the mass of 1 mole of a substance, obtained by summing the masses of the component atoms

26. Chemical Composition Calculate the molar mass of sulfur dioxide, a gas produced when sulfur containing fuels are burned. Formula SO 2 How many S atoms in 1 molecule? 1 How many O atoms in 1 molecule? 2 1(32.07g) + 2(15.999g)= 64.07g

27. Chemical Composition Calcium carbonate, CaCO 3 (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?

28. Chemical Composition 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 C 10 H 6 O 3. 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?

29. Chemical Composition Isopentyl acetate, C 7 H 14 O 2 , 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 -6 g. How many moles and how many molecules of the isopentyl acetate are released in a typical bee sting?

30. Chemical Composition 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

31. Chemical Composition Given the compound C 2 H 5 OH 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

32. Chemical Composition 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

33. Chemical Composition Carvone is a substance that occurs in two forms, both of which have the same molecular formula (C 10 H 14 O) 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.

34. Chemical Composition 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

35. Chemical Composition Determining Empirical Formulas: C 6 H 6 CH C 12 H 4 Cl 4 O 2 C 6 H 2 Cl 2 O C 6 H 16 N 2 C 3 H 8 N

36. Chemical Composition 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

37. Chemical Composition 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.

38. Chemical Composition 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.

39. Chemical Composition 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.

40. Chemical Composition 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.

41. Chemical Composition A white powder is analyzed and found to have an empirical formula of P 2 O 5 . The compound has a molar mass of 283.88. What is the compound’s molecular formula?