# The Mole Chemical Formulas The Mole n Mole

- Slides: 15

The Mole & Chemical Formulas

The Mole n Mole, abbreviated mol, is the SI (metric) base unit to measure the amount of a substance – It is called Avogadro’s number in honor of Italian physicist Amedeo Avogadro, who discovered one mole of gas in 1811 – 6. 0221367 x 1023 representative particles 602, 000, 000, 000

Use of the Mole n One wouldn’t use this unit for everything – A mole of marbles would cover the surface of the Earth down to approximately 6 kilometers – Although a mole of $1. 00 bills would be nice! n It is used to measure the amount of molecules in a substance

Conversion Recall n 5, 280 feet = 1 mile n How many miles would you have in 12, 369 feet? n It is the same system of proportions except now its… – 1 mole = 6. 02 x 1023 – So, if you had 2. 5 moles of H 2 O, how many atoms of water do you have?

Recall n Determine the number of atoms in 2. 75 moles of Zn? n If you had 9 moles of Cu(OH), how many molecules are present? n Lets change it up! – How many moles of H 2 SO 4 if you had 2. 11 x 1024 molecules?

Mass & The Moles n A dozen oranges won’t have the same mass a dozen watermelons – But you do have exactly the same amount of each § 12 n Since they have different compositions, their mass will be different as well – Thus, the mass of 1 mole of silver nitrate (Ag. NO 3) will be different than 1 mole of oxygen gas (O 2)

Periodic Table Recall n The atomic masses on the periodic table are not whole numbers – They are weighted averages of the masses of all naturally occurring isotopes of each element n Knowing this and using this new idea… – Carbon – 12 is the standard atom used to determine the number of representative particles per mole § We can calculate the molar mass or mass in grams of one mole of any pure substance (g/mol)!

So What Does This Mean? n Periodic Table Transitions** – Carbon has a mass of 12 § Thus 12 g of carbon = 1 mole of Carbon – Manganese (Mn) has an atomic mass of 54. 94 amu § 55 g of Mn = 1 mol Mn n n You Try! – – – 3 mol Mn = ? g Mn 136 g Pb = ? Mol Pb Challenge* § How many grams of gold are found in 7. 65 x 1022 Au atoms? This works for compounds as well as single elements – 1. 00 g H 2 O = ? Mol H 2 O – 1. 00 g H 2 O = ? Molecules H 2 O

Percent Composition n Percent Composition: The percentage by mass of each element in a compound – It is determined by gravimetric and volumetric analyses for solids and liquids, respectively Example: 100 g of an unknown substance has two elements: X & Y X has 55 g and y has 45 g of the total 100 (55 g of element X / 100 g cmpd. )*100% = 55% (45 g of element Y / 100 g cmpd. )*100% = 45%

Percent Composition n If you know the compound in question, you can calculate the percent composition for the known atoms Example: 18. 02 g/mol H 2 O Hydrogen (H) – 1. 00794 g Oxygen (O) – 15. 9994 g 1. 00794 g * 2 Hydrogen in water = 2. 02 g (2. 02 g H / 18. 02 g H 2 O) * 100 = 11. 2 % H 16. 00 g O * 1 Oxygen in water = 16. 00 g (16. 00 g O / 18. 02 g H 2 O) * 100 = 88. 8 % O So, water is 11. 2% Hydrogen and 88. 8% Oxygen Try 8. 00 g of Cu(OH)

Empirical Formula n After you have identified the compound and its percent composition, you can find the formula of the compound – First, you must find the smallest whole number ratio of the moles of the elements in the compound § It provides the subscripts for the chemical formula n Empirical formula is the lowest whole number ratio of the elements – CH 2 O … C 2 H 4 O 2 … C 3 H 6 O 3 … C 4 H 8 O 4 … C 5 H 10 O 5 … etc. MOST IMPORTANT FACT * Empirical formula DOES NOT have to match actual molecular formula Example: HO is the empirical formula for hydrogen peroxide, H 2 O 2, but HO is not hydrogen peroxide!

Empirical Formula n You can assume the mass of the compound is 100 g if they give you percent composition. So if 40. 05 % is Sulfur (S) and 59. 95% is Oxygen (O), then there is 40. 05 g S and 59. 95 g. O Then simply convert to moles: 40. 05 g S * 1 mol S = 1. 249 mol S 1 32. 07 g S 59. 95 g O * 1 mol O 1 16 g O = 3. 747 mol O So the molar ratio is 1. 249: 3. 747 for S: O atoms This will produce integers but we want whole numbers How do we simplify? Take the smaller number and divide it and everything else by that number. 1. 249/1. 249 = 1 3. 747/1. 249 = 3 1: 3 Ratio so for every one S, there is three O. SO 3

Practice Problems 1. Ni: 34. 4%, S: 37. 5%, O: 28. 1% 2. C: 69. 7%, H: 14. 0%, N: 16. 3% 3. C: 60. 6%, H: 8. 14%, P: 31. 3%

Molecular Formula n Knowing what you know about building compounds, percent composition and empirical formulas… – Realize that hundreds of different substance have the same percent composition and empirical formula n It is possible because they have different molecular formulas – This specifies the actual number of atoms of each element in one molecule or formula unit of the substance

Molecular Formula n To determine the molecular formula, the molar mass of the compound must be determined through experimentation and compared with mass represented by the empirical formula – Example: Acetone has a molar mass of 26. 04 g/mol while its empirical formula is CH or 13. 02 g/mol § Since the mass of the molecule is 26. 04 g/mol, divide it by the empirical mass of 13. 02 g/mol – 26. 04 g/mol / 13. 02 g/mol = 2. 00 – Thus the molecular formula has to have twice as many carbon and hydrogen atoms than the empirical formula, or C 2 H 2 n (Empirical Formula)*n = Molecular Formula

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