Week 4 Chemistry Notes
Week 4 Chemistry Notes CHEM 1127Q-011
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This 4 page Class Notes was uploaded by Amelia on Wednesday September 21, 2016. The Class Notes belongs to CHEM 1127Q-011 at University of Connecticut taught by Dr. Cady in Fall 2016. Since its upload, it has received 18 views. For similar materials see General Chemistry in Chemistry at University of Connecticut.
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Date Created: 09/21/16
9/19 *Notes from last chapter: When naming Binary Ions, Zinc and Silver are the only transition metals that do not need Roman Numerals, because Zinc is always Zn and Silver is always Ag . Also, Aluminum is always Al . + *We are devoting so much time to this chapter because naming is a good portion of the first test. CHAPTER 3 Compositions of Substances and Solutions Ratios and Stoichiometry 1) Na P3 : 4ams are no use for comparison in elements 2) Moles and the number of atoms are the key to relating products to reactants, we need numbers for things The Mole 1) How much is a mole? a) A mole of pennies would be a stack 300 meters tall. 2) 1.00 mol of atoms is 6.022 × 10 23 a) Known as Avogadro's number 3) Mass = MM x n a) MM= molar mass (g/mole) b) n= number of moles 4) Examples a) NaCl i) 1(22.99g/mole) + 1(35.45g/mole) = 58.44g/mole b) If you have 0.250g how many moles of Sucrose do you have? i) Molecular Formula: C H O12 221111) + 22(1.008) + 11(16.00)= 342.30g/mole ii) 0.250g ÷ 342.30g/mole = 0.000730 moles c) How many moles of Iron(III) oxide in 3g? i) Iron (III) oxide = Fe O 2 3 ii) 2(55.85g/mole) + 3(16.00g/mole) = 159.7g/mole iii) 3g ÷ 159.7 = 0.0188moles d) How many molecules in 3g of Iron (III) oxide? i) 0.0188moles × 6.022×10 = 1.132×10 molecules 22 5) Converting between molecules and component atoms a) For every one molecule of FE O the2 3e 3 oxygen atoms b) Just like a car has 4 tires or a chair has four legs c) Example: i) How many oxygen atoms are in 3 grams of Fe O ? 2 3 ii) 1.132×10 molecules Fe O × 3 O2g3÷ 1 Fe O 2 3 Mass Percent of Compounds 1) Sometimes you want to know what percent of your sample is a certain element MassX 2) Mass Percentage= ×100 Mass of Sample 3) Example: a) Mass percent of Fe in 3g Fe O 2 3 b) 3g ÷ 159.7g/mole = 0.0188 moles Fe O × 2Fe÷1Fe O = 0.0376moles of Fe 2 3 2 3 c) 0.0376 Fe × 55.85g/mole = 2.10g Fe d) (2.10g Fe÷ 3g Fe O ) × 100 = 70% 2 3 4) Mass percent of Sulfate in 20g Ammonium Sulfate a) (NH ) SO → SO 2= 96.06g/mole, NH = 18.042g/mole 4 2 4 4 4 b) (NH ) SO = 132.14g/mole 4 2 4 c) SO 2= (96.06g/mole÷ 132.14g/mole) × 100 = 72.7% 4 5) You have an 20g of an unknown sample. 15g of the sample is Sulfate. What is the mass percent of the Cation? a) Cation(+) + Anion() = Total 20g b) X + 15g = 20g c) X=5g 5g d) × 100 = 25% 20g 9/21 6) 4.319g of dimethyl nitrosamine (containing C, H, N, and O) burned in oxygen yields 5.134 carbon dioxide and 3.173g of water. The compound contains 37.82% by mass of nitrogen. What is the empirical formula of dimethyl nitrosamine? a) Carbon dioxide → CO Water → H2 Nitrogen→ N 2 i) Oxygen can also come from air, so cannot be assumed. So we need to find it as the leftover. b) CO → 214g÷ 44.01g/mole = 0.1166 moles of CO × 1C/1CO = 0.11662oles Carbon 2 c) H O 23.173 ÷ 18g/mole = 0.1763 moles of H O × 2H/1H O = 0.326 moles Hydroge2 d) N → 37.82% … 37.82× 4.319 = 1.633g N ÷ 14.01g/mole = 0.1166 moles of Nitrogen e) O → ? i) Total mass = mass Carbon + mass Hydrogen + mass Nitrogen + mass Oxygen ii) 4.319g = (0.1166moles×12.01g/mole) + (0.3526 moles×1.008g/mole) + (0.1166 moles ×14.01g/mole) iii) 4.319g = 1.40g + 0.355g + 1.630g iv) Oxygen = 0.931g÷ 16.00g/mole = 0.0582 f) Moles of elements i) C = 0.1166moles ii) N = 0.1166moles iii) H = 0.355moles iv) O = 0.0582moles g) Divide everything by smallest moles (in this case oxygen) i) C = 2 ii) N = 2 iii) H = 6 iv) O = 1 h) C H N O 2 6 2 Molarity 1) We often talk about things in terms of concentration moles/Liter 2) Whenever you see [x] square brackets think concentration 3) Molarity is also represented by M which stands for moles/Liter 4) Abbreviations a) mm= molar mass b) M = moles per Liter c) N = Number of Moles 5) Example a) 19.4g of Potassium Chromate dissolved in 1 Liter. What is the molarity? b) K C2 = 44.20g/mole c) 19.4g ÷194.20g/mole = 0.100 moles d) 0.100 moles/1 liter = 0.1M 6) What is [K ]? + + a) 0.100 moles K CRO 22 = 0400 moles K + b) 0.200 moles K /1 Liter = 0.2M c) Remember your ionic formulas 7) How much 0.100 M CaCl solution a2 water must be mixed together to make 2.00 liters of 0.0250M Cl solution? a) Goal: 2.00 liters of 0.0250M or 0.05 moles 2.00 Cl b) Start: 0.100M CaCl ×(2Cl ÷ 1 CaCl ) = 0.200M Cl 2 2 c) 0.05 Cl ÷0.200 mole/Liter Cl = 0.25L d) 0.25L + 1.75L H O = 2.00L 2 Percentages MassX 1) Mass Percentage= Mass of Sample ×100 2) Volume Percentage= VolumeX ×100 Volume of Sample 3) ppm = MassX × 10 Total mass 4) ppb= MassX × 10 9 Total mass 5) Example a) Your water is tested and found to have 175 ppm CaCo . What is the molarity of Ca ? 2+ 3 Assume the density of the solution is 1g/ml. b) ppm = MassX × 10 Total mass c) 175 ppm = x × 10 1000g 6 x d) 175 × 10 = 1000g e) x = 0.175g CaCO 3 f) 0.175g ÷100.0g/mole = 0.00175 moles CaCO × 1Ca/1 CaC3 = 0.00175 moles/Liter g) 0.00175M No prelab this week because next week we use our lab time to study for the first exam! I’ll put out a study guide Friday, and the test is a week after. Good luck!
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