Chem 1211 1st Edition Lecture 6 Outline of Last Lecture I. Atomic NumbersII. IsotopesIII. Law of Definite ProportionsIV. Ions and Ionic CompoundsV. Complex IonsVI. MolesOutline of Current Lecture I. Percent CompositionII. Formulas from Elemental Composition Current LectureI. Percent Composition (elemental analysis)a. Comes from experiments b. Gives all the elements and what percentage they occur ini. Example- In CH4, what is the percent Hydrogen by mass and the percent Carbon by mass?1. There is 1 Carbon at 12.01g2. There are 4 hydrogens at 1.008 ga. 12.01 + 4.032 gives a total mass of 16.042b. (12.01/16.042) = percent carbon = 75 %c. (4.032/16.042) = percent hydrogen = 25 %ii. OR… we have an unknown compound, and the only information we have is that it is 25% H (mass) and 75 % C (mass).1. We say that we have 100 g of the sample (easier math)a. (0.25 g H/ g substance) x 100 = 25 g H x (1mol/1.008 g) = 25 mol HThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.b. (.75 g C/ g substance) x 100 = 75 g C x (1mol/12.01) = 6.2 mol Cc. (25 mol H/6.2 mol C)= 4 H per each Cd. CH42. Take % find grams of each element moles of each element c. Concept Test: Which molecules have the lowest and highest percent compositionof oxygen by mass, respectively…i. Acetic acid, CH3CO2Hii. Carbon monoxide, COiii. Ethylene glycol, HOCH2CH2OH1. ANSWER: ethylene glycol has the lowest percent composition of oxygen by mass and carbon dioxide has the highest percent composition of oxygen by massa. Reasoning: the numerator is the amount of oxygen in the molecule, in gramsb. The denominator is the total amount of the molecule, in gramsc. Acetic = (32/60)= 53 %d. CO = (16/28) = 57%e. Ethylene = (32/62)= 52 %d. Ratiosi. If you find that the ratio of A to B is 1.5: 1, there must be 3 moles of A to every 2 moles of B1. You need an integerII. Formulas from elemental composition a. Empirical formula – the smallest whole number ratioi. Ex. CH2 b. Molecular formula i. Ex C2H4c. To go from empirical molecular formula, you need a molecular weighti. C2H4 CH21. Molecular weight C2H4 = 28 ii. C5H10 CH2 1. Molecular weight C2H4 = 70iii. The molecular weight tells you how many times the empirical formula appears in the molecular formula d. Law of Multiple Proportions – one of two things must be true; different molecules have different ratios of elements, even if they have the same elements(aka water and hydrogen peroxide); or, if they have the same elemental formulas,those formulas represent different structuresi. Restated: when two elements can combine to form more than one compound the amounts of one of them that combines with a fixed amount of the other will exhibit a simple multiple relatione. Concept Test: Two organic molecules are dimethyl ether, CH3OCH3 and ethanol CH3CH2OH. Which is the most accurate statement about the two molecules?i. Obey Law of Multiple Proportions because they have different structural formulasii. Violate Law of Multiple Proportions because they have identical elemental formulas, C2H6Oiii. Are not subject to the Law of Multiple Proportions because they are organiciv. This is a trick question, so the formulas are not drawn accurately 1. ANSWER: i (they have different structural formulas)f. Concept Test: A Rolaids antacid has 550 mg of calcium carbonate and 110 mg magnesium hydroxide per tablet. What is the percentage of calcium by mass per tablet? (Mg= 24.3, Ca= 40.8, HO=17.01, CO3= 60.01)i. ANSWER: (220 mg/ 660 mg) = 33.3 %1. Reasoning: denominator is the total mass of the tablet (550+110)2. The numerator is the mass of Calcium in the tablet. Since Calcium makes up 40% of the 550 mg of calcium carbonate (CaCO3), 0.40 x 550 = 220g. Percent Purityi. Percent purity = (mass of pure substance/mass of sample) x
View Full Document
Unlocking...