Unit 2 Formulas and Equations AP Chemistry Unit 2 Formulas and Equations

The mole is the unit for counting atoms and molecules 6.02 × 1023 atoms of (insert name of element here) has a mass of (insert mass of element from periodic table here)

How many atoms are in 5.00 g of oxygen gas 5.00 g O 2 1 × 1 mol O 2 32.0 g O 2 × 2 mol O 1 mol O 2 × 6.02 × 10 23 atoms 1 mol O = 1.88 × 1023

3.64 g NaCl 50.0 mL × 1 mol NaCl 58.5 g NaCl × 1000 mL 1 L =1.24 M Determine the concentration (in molarity) when 3.64 g of NaCl is added to enough water to make 50.0 mL of solution. 3.64 g NaCl 50.0 mL × 1 mol NaCl 58.5 g NaCl × 1000 mL 1 L =1.24 M

3.19 mmol Ca NO 3 2 25.0 mL × 1 𝑚𝑜𝑙 1000 𝑚𝑚𝑜𝑙 × 1000 𝑚𝐿 1 𝐿 =0.128 𝑀 Determine the concentration (in molarity) when 3.19 mmol of Ca(NO3)2 is added to enough water to make 25.0 mL of solution. 3.19 mmol Ca NO 3 2 25.0 mL × 1 𝑚𝑜𝑙 1000 𝑚𝑚𝑜𝑙 × 1000 𝑚𝐿 1 𝐿 =0.128 𝑀 Note that the milli- in mmol and mL effectively canceled!

A student has 12. 0 M HCl available but needs 50. 0 mL of 0 A student has 12.0 M HCl available but needs 50.0 mL of 0.500 M for an experiment. How much of the concentrated acid should she use? 0.500 mol HCl 1 L HCl (𝑑) × 0.0500 L HCl (𝑑) 1 × 1 L conc. HCl (𝑠) 12.0 mol HCl 0.00208 L or 2.08 mL

Recognizing that molarity times volume gives moles… 20.0 mL of 0.280 M HNO3 is diluted to 100.0 mL. Determine the new concentration of nitric acid Recognizing that molarity times volume gives moles… 0.280 mol H NO 3 1 L × 20.0 mL 100.0 mL =0.00560 M

Composition stoichiometry deals with the makeup of a single compound How many grams of iron is in 3.85 g of Fe2(SO4)3? 3.85 g Fe 2 SO 4 3 1 × 1 mol Fe 2 SO 4 3 399.9 g Fe 2 SO 4 3 × 2 mol Fe 1 mol Fe 2 SO 4 3 × 55.8 g Fe 1 mol Fe =1.07 g

F.M. of CuCO3 is 187.0 g Mass of copper is 63.5 × 2 = 127.0 g What percent of Cu2CO3 is copper? F.M. of CuCO3 is 187.0 g Mass of copper is 63.5 × 2 = 127.0 g %Cu= 127.0 g Cu 187.0 g Cu 2 CO 3 ×100=67.9%

A balanced chemical equation is necessary All problems follow Reaction stoichiometry deals with relationships in chemical reactions A balanced chemical equation is necessary All problems follow g A → mol A → mol B → g B All fractions, other than the first (given), must equal one!

2 C 4 H 10 +13 O 2 →10 H 2 O+8 CO 2 Butane burns in oxygen according to the equation shown above. When 15.85 g of butane burns in excess oxygen, how much water should be formed? 15.85 g C 4 H 10 1 × 1 mol C 4 H 10 58.0 g C 4 H 10 × 10 mol H 2 O 2 mol C 4 H 10 × 18.0 g H 2 O 1 mol H 2 O =24.6 g

We have 11.85 g of Na 3 PO 4 and we need 11.61 g 3Ca NO 3 2 +2 Na 3 PO 4 → Ca 3 PO 4 2 +6 NaNO 3 17.42 g of calcium nitrate reacts with 11.85 g of sodium phosphate according to the reaction shown above. Determine which one is the limiting reactant. 17.42 g Ca NO 3 2 1 × 1 mol Ca NO 3 2 164.1 g Ca NO 3 2 × 2 mol Na 3 PO 4 3 mol Ca NO 3 2 × 164.0 g Na 3 PO 4 1 mol Na 3 PO 4 = 11.61 g Na 3 PO 4 We have 11.85 g of Na 3 PO 4 and we need 11.61 g Therefore Na 3 PO 4 is the excess reactant Which means Ca NO 3 2 is the limiting reactant!

Fe 2 O 3 +2Al→ Al 2 O 3 +2Fe 12.91 g of iron(III) oxide reacts with aluminum as shown above. If 8.87 g of molten iron is formed, what is the percent yield? 12.91 g Fe 2 O 3 1 × 1 mol Fe 2 O 3 159.6 g Fe 2 O 3 × 2 mol Fe 1 mol Fe 2 O 3 × 55.8 g Fe 1 mol Fe =9.03 g Fe %𝑌𝑖𝑒𝑙𝑑= 8.87 9.03 ×100=98.2%

2) determine moles of each element An empirical formula is a chemical formula in lowest terms. Steps in determining an empirical formula from percent composition. 1) assume 100 g if given in % 2) determine moles of each element 3) divide each molar amount by the smallest one 4) check for ½’s 1/3’s ¼’s etc.

Determine the empirical formula for a compound that is found to be 52 Determine the empirical formula for a compound that is found to be 52.14% C, 13.13% H and 34.73% O C = 52.14 g H = 13.13 g O = 34.73 g C = 52.14 g / 12.011 g/mol = 4.341 mol H = 13.13 g / 1.0079 g/mol = 13.027 mol O = 34.73 g /16.00 g/mol = 2.171 mol Divide all by the smallest, 2.171 mol C = 4.341 mol / 2.171 mol = 1.9999 H = 13.027 mol / 2.171 mol = 6.000 O = 2.171 mol / 2.171 mol = 1.000 E.F. = C2H6O

Determine the empirical formula for a compound that is found to be 24 Determine the empirical formula for a compound that is found to be 24.52 % S, 26.52% Cr, and 48.96% O S = 24.52 g Cr = 26.52 g O = 48.96 g S = 24.52 g / 32.06 g/mol = 0.7648 mol Cr = 26.52 g / 52.00 g/mol = 0.5100 mol O = 48.96 g / 16.00 g/mol = 3.060 mol Divide all by the smallest, 0.5100 moles S = 0.7648 mol / 0.5100 mol = 1.4996 Cr = 0.5100 mol / 0.5100 mol = 1.000 O = 3.060 mol / 0.5100 mol = 6.000 E.F. = Cr2S3O12 or Cr2(SO4)3

A molecular formula can be determined from an empirical formula if additional information is given, such as molar mass.

Determine the molecular formula for a compound that is found to be 9 Determine the molecular formula for a compound that is found to be 9.15% hydrogen, 36.32% oxygen, and 54.53% carbon. The molar mass was found to be 120 g/mol +/-25 g/mol. C = 54.53 g H = 9.15 g O = 36.32 g C = 54.53 g / 12.011 g/mol = 4.540 mol H = 9.15 g / 1.0079 g/mol = 9.078 mol O = 36.32 g /16.00 g/mol = 2.270 mol Divide all by the smallest, 2.270 mol C = 4.540 mol / 2.270 mol = 2.000 H = 9.078 mol / 2.270 mol = 3.999 O = 2.270 mol / 2.270 mol = 1.000 E.F. = C2H4O

Determine the molecular formula for a compound that is found to be 9 Determine the molecular formula for a compound that is found to be 9.15% hydrogen, 36.32% oxygen, and 54.53% carbon. The molar mass was found to be 120 g/mol +/-25 g/mol. E.F. = C2H4O Mass of E.F. is 2 × 12.0 4 × 1.0 1 × 16.0 44.0 g/mol 1 × E.F. = 44.0 g/mol 2 × E.F. = 88.0 g/mol 3 × E.F. = 132.0 g/mol 4 × E.F. = 176.0 g/mol 5 × E.F. = 220.0 g/mol Range is 95 – 145 g/mol M.F. = 3 × E.F. C6H12O3

Many ionic substances that crystallize in water trap water molecules in the structure. These “hydrates” have very specific formulas For example: MgSO4·7H2O

A hydrate of magnesium chloride is placed into a crucible of mass 22 A hydrate of magnesium chloride is placed into a crucible of mass 22.130 g. Before heating the crucible and contents have a mass of 25.290 g. After heating the mass is 23.491 g. Determine the formula for the hydrate. Mass of MgCl2 is 23.491 – 22.130 = 1.361 g Mass of H2O is 25.290 – 23.491 = 1.799 g Moles of MgCl2 is 1.361 / 95.21 = 0.01429 mol Moles of H2O is 1.799 / 18.02 = 0.09983 mol Dividing by 0.01429 gives MgCl2·7H2O

Hydrates MgCO3∙5H2O Na2CO3∙2H2O Na2CO3∙10H2O 2 10 LiClO4∙3H2O ZnSO4∙7H2O BaCl2∙2H2O CoCl2∙2H2O

Oxidation Numbers Ion’s charges are their oxidation numbers The most electronegative element (usually the last one) is treated like an ion Everything else must be calculated

Determine the oxidation number of chlorine in chlorate ClO3- Cl = x O = -2 × 3 = -6 x + (-6) = -1 x = 5

Determine the oxidation number of chromium in Al2(Cr2O7)3 Cr = 6x O = -2 × 21 = -42 6 + 6x + (-42) = 0 x = 6 or Cr2O72- Cr = 2x O = -2 × 7 = -14 2x + (-14) = -2 x = 6

Strong acids ionize completely in water There are seven strong acids They are formed when hydrogen bonds to soluble anions NO3- ClO4- ClO3- Cl- Br- I- SO42- The only exception is C2H3O2-

C2H4Cl2 can be put together in two different ways Isomers exist among some organic compounds. They have the same formula but a different structure C2H4Cl2 can be put together in two different ways H− Cl | C | H − Cl | C | H −H or Cl H C=C Cl H or H− Cl | C | Cl − H | C | H −H Structural isomers Cl H C=C H Cl Optical isomers

Any carbon with four different things bonded to it will give rise to an isomer Cl− H | C | Br −I vs Cl− I | C | Br −H

Most metals oxidize but in different ways Iron will oxidize (rust) until the whole sample is consumed Aluminum forms an oxide layer (tarnish) that protects the underlying metal Stainless steel contains chromium which forms a protective oxide layer to keep the iron from oxidizing Steel can be coated with a protective layer of zinc in a process called galvanization

Spectrophotometry uses light to determine the concentration of colored solutions

Absorbance varies with the wavelength of light used

Once the optimal wavelength is chosen, a series of standard solutions is prepared and tested

Spectrophotometry makes use of the Beer-Lambert law

It boils down to one equation: A = abc or A = εlc Where: A is the absorption of light a (ε) is the molar absorptivity b (l) is the path length c is the concentration A = -log T 𝑇= 𝐼 𝑡 𝐼 0 Where: T is transmittance It = Intensity of light transmitted I0 = Intensity of original light

Quiz 1) 3.42 × 1024 9) C9H20 2) 1.11 × 1025 10) 22.508 g 3) 2.66× 1020 11) Ni3(PO4)2·12H2O 4) 2.73× 1024 12) CaCl2·2H2O 5) 0.00239 M 13) (a) 5 6 (a) 0.02974 M (b) 1 (b) 0.08923 M (c) -½ (c) 0.1190 M (d) 6 7) 30.34% 14) 0.00062 M 8) 97.4% 15) 0.49 M