# Moles and Atoms

Counting Atoms by the Gram

- Mole
- this is the SI base unit for the amount of substance
- by knowing this, you can calculate the amount of particles of a substance, the number of grams of a substance, and calculate the volume of a gas at a given temperature and pressure
- 1 mole = 6.022 x 10^23 particles
- avogadro's number
- two main substances are ionic compound and molecular compounds (you also have elements)
- You have acids if they have an H in front and these types of substances are types of molecular compounds
- element - atoms
- molecular compounds - molecules
- ionic compounds- formula units
- Problems
- Convert 2.85 mol of He to atoms
- 2.85 mol/1 * 6.022 x 10^23 atoms/1mol = 1.72 x 10^24 atoms of Helium

- Convert 8.57 x 10^23 atoms of Cu to moles
- 8.57 x 10^23 atoms/1 * 1 mol/6.022 x 10^22 atoms = 1.42 moles of Copper

- Convert 4.52 mol of Na to atoms
- 4.52 mol/1 * 6.022 x 10^23 atoms/1mol = 2.72 x 10^24 atoms of Sodium

- Convert 7.94 x 10^22 atoms of S to moles
- 7.94 x 10^22 atoms/1 * 1mol/6.022 x 10^23 atoms = 0.132 moles of Sulfur

- molar mass
- the sum of all of the molar masses of the atoms of that particular formula
- periodic table - round to the second decimal place
- molar mass = 1 mole = 6.022 x 10^23 particles

- He
- 1 mole
- 4.00g
- 6.022 x 10^23 atoms
- CO2

- 1 mole
- 44.01g
- 6.022 x 10^23 molecules

- CuSO4
- 1 mole
- 159.62g
- 6.022 x 10^23 formula units

- Problems
- Convert 43.76 g of Br to moles
- 43.76g/1 * 1mol/79.90g = .5477 moles of Bromine

- Convert 3.76 mol of neon to mass (grams)
- 3.67 mol/1 * 20.18g/1mol = 74.1 g of Neon

- Convert 2.35 x 10^23 atoms of zinc to mass (grams)
- 2.35 x 10^23 atoms/1 * 1mol/6.022×10^23 atoms * 65.39g/1mol = 25.5g of Zinc

- Convert 6.89g of Ca to atoms
- 6.89g/1 * 1mol/40.08g * 6.022 x 10^23 atoms/1mol = 1.04 x 1023 atoms of Calcium

- Problems
- Convert 15.6 moles of H2) to molecules
- 15.6 mol/1 * 6.022 x 10^23/1mol = 9.39 x 10^24 moles of water

- Convert 7.83 x 10^22 formula units of Copper(II) chloride to moles
- 7.83 x 10^22 formula units CuCl2/1 * 1mol CuCl2/6.022 x 10^23 formula units CuCl2 = 0.130 moles of CuCl2

- Convert 6.72 moles of carbon monoxide to mass(g)
- 6.72 moles CO/1 * 28.01g/1mol CO = 188 g of CO
- 25.8g of calcium hydroxide to moles
- 25.8g Ca(OH)2/1 * 1mol Ca(OH)2/74.10g = 0.348 moles of Ca(OH)2

- Convert 34.82 g of nitrogen dioxide to molecules
- 48.82g NO2/1 * 1mole NO2/46.01 gNO2 * 6.022 x 10^23 molecules NO2/1mol = 4.557 x 10^23 molecules of NO2

- Convert 7.83 x 10^24 formula units of lithium carbonate to grams
- 7.83 x 10^24 fu Li2CO3/1 * 1mol LI2CO3/6.022×10^23 fu Li2CO3 * 73.89g/1mol Li2CO3 = 961 g of Li2CO3

- Convert 8.63 x 10^23 molecules of CH3OH convert to mass
- 8.63 x 10^23 molec CH3OH * 1 mol CH3OH/6.022 x 10^23 molec CH3OH * 32.05g CH3OH/1 mol CH3OH = 45.9 g of CH3OH

- Convert 23.87 g of silver nitrate to particles
- 23.87g AgNO3/1 * 1mol AgNO3/169.88 g AgNO3 * 6.022 x 10^23 particles AgNO3/1mol AgNO3 = 8.462 x 10^22 particles of AgNO3

- Determine the number of moles of O in 8.54 moles of nitrogen trioxide
- 8.54 mol NO3 * 3 mol O/1mol NO3 = 25.6 mol of O

- Determine the number of atoms of N in 9.32 x 10^24 formula units iron (II) nitrate
- 9.34 x 10^24 form units Fe(NO3)2 * 2 atoms N/1 form unit Fe(NO3)2 = 1.87 x 10^25 atoms of N

- Calculate the grams of C in 45.78 g of C2H5OH
- 45.78g C2H5OH/1 * 1 mol C2H5OH/46.08 g C2H5OH * 2 mol C/1 mol C2H5OH * 21.01 g C/1 mol C = 23.86 g of C

- Calculate the grams of N in 52.3 g of nitrogen trichloride
- 52.3 g NCl3/1 * 1 mol NCl3/120.36 g NCl3 * 1 mol N/1 mol NCl3 * 14.01 g N/1 mol N = 6.09 g N

- Calculate the number of hydrogen atoms in 15.8 g of CH4
- 15.8 g CH4/1 * 1 mol CH4/16.05 g CH4 * 4 mol H/1 mol CH4 * 6.022 x 10^23 atoms H /1 mol H = 2.37 x 10^24 atoms of H

- Mass Percent Composition
- mass percent composition (mass percent; percent composition)
- Total mass of the element/total mass of the compound x 100%
- 2 ways to determine
- experimental evidence
- chemical formula

- Ammonium nitrate - NH4NO3
- molar mass = 80.05 g/mol
- 1 mole (80.05g) contains 28.02g of nitrogen
- Percent N = 2802g/80.05g x 100% = 35.00%
- Always round percent by mass to the first decimal place
- Problems
- What is the percent composition of a compound formed when 2.70g of aluminum combine with oxygen to form 5.10g of aluminum oxide?
- Al% = 2.70g/5.10g x 100% = 52.9%
- O% = 47.1%
- Calculate the percent composition when 13.3g Fe combine completely with 5.7 g of O.
- Fe% = 13.3g/19g = 70.0%
- O% = 30.0%
- Determine the percent composition of magnesium cyanide
- Mg% = 24.31g/76.35g x 100 = 31.8%
- C% = 24.02g/76.35g x 100 = 31.5%
- N% = 36.7%
- Determine the percent composition of diphosphorus trioxide.
- P% = 62.94g/109.94 = 56.3%
- O% = 43.7%
- The FDA recommends that adults consume less than 2.4g of sodium per day. How many grams of sodium chloride can you consume and still be within the GDA guidelines?

- 2.4g Na/1 * 1mol Na/22.99 g Na * 1 mol NaCl/1mol Na * 58.44 g NaCl/1mol NaCl = 6.1g NaCl
- Calculating Empirical Formulas
- Empirical formula
- Molecular formula
- C4H10, C6H12, H2O2, C3H8
- Steps for solving empirical formula
- convert to moles
- divide each amount of moles by the lowest number of moles
- use the ratio between the moles to develop the empirical formula
- Steps for solving the molecular formula
- calculate the empirical formula
- molecular formula = molar mass/empirical mass = integer
- use the integer to multiply the subscripts of empirical formula

- Problems
- Determine the empirical and molecular formula if a compound contains 71.72% Cl, 16.16% O, and 12.12% C; molar mass = 198g
- 71.72 g Cl/1 * 1 mol Cl/35.45g Cl = 2.02 mol Cl
- 16.16 g O/1 * 1 mol O/16.00 g O = 1.01 mol O
- 12.12 g C/1 * 1 mol C/1201 g C = 1.01 mol C
- 2.02 mol Cl/1.01 mol O = 2 mol Cl/1 mol O
- 1.01 mol C/1.01 mol O = 1 mol C/1 mol O

- COCl2
- MF = MM/EM = 198g/98.9g = 2
- C2O2Cl4
- Determine the empirical and molecular formula if a compound contains 94.1% O and 5.9% H; molar mass = 34g
- 94.1g O * 1mol O/16 g O = 5.88 mol O
- 5.9g H * 1mol H/1.01g H = 5.84 mol H
- 5.84 mol H/5.88 mol O = 0.993 mol H/1 mol O
- HO
- MF = MM/EM = 34/17.01 = 2.0
- H2O2
- Determine the empirical and molecular formula if a compound contains 40.05% C, 6.6% H, and 53.4% O; molar mass = 120g.
- 40.05 g C * 1 mol C/12.01 g C = 3.34 mol C
- 6.6 g H * 1 mol H/1.01 g H = 6.53 mol H
- 53.4 g O * 1 mol O/16.00 g O = 3.34 mol O
- 3.34 mol C/3.34 mol O = 1mol C/1mol O
- 6.53 mol H/3.34 mol O = 1.96 mol H/1mol O
- H2CO
- MF = MM/EM = 120/30.03 = 4.0
- H8C4O4