Stoichiometry Part 1 PDF

Part 1 The terms "mass" and "weight" are used interchangeably in ordinary conversation, but the two words don't mean the same thing. The difference between mass and weight is that mass is the amount of matter in a material, while weight is a measure of how the force of gravity acts upon that mass....

Part 1 The terms "mass" and "weight" are used interchangeably in ordinary conversation, but the two words don't mean the same thing. The difference between mass and weight is that mass is the amount of matter in a material, while weight is a measure of how the force of gravity acts upon that mass. Mass is the measure of the amount of matter in a body. Weight is the measure of the amount of force acting on a mass due to the acceleration due to gravity. Weight usually is denoted by W. Weight is mass multiplied by the acceleration of gravity (g). W=mxg Example. Calculate the molecular mass of ethanol, whose condensed structural in CH3CH2OH. Among its many uses, ethanol is a fuel for internal combustion engines. Given: molecules Asked for: molecular mass Solution: A. Determine the number of atoms of each element in the molecule. The molecular formula of ethanol may be written in three different ways: CH3CH2OH (which illustrates the present of an ethyl group, CH3CH2 －, and an －OH group), C2H5OH and C 2 H 6 O; all show that ethanol has two carbon atoms, six hydrogen atoms and one oxygen atom. B. Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each elements by the number of atoms of that element. 2 x atomic mass of carbon = 2 x 12 g/mole = 24 g/mole 6 x atomic mass of hydrogen= 6 x 1 g/mole = + 6 g/mole 1 x atomic mass of oxygen = 1 x 16 g/mole = + 16 g/mole 46 g/mole C. Add together the masses Note: for convention atomic masses of element should be round off to the nearest tenths or hundredths. Example atomic mass of oxygen is 15.9994 g/mole converted to 16 g/mole; hydrogen atomic mass is 1.0079 g/mole converted to 1 g/mole; and carbon atomic mass 12.011 g/mole converted to 12 g/mole. Try to solve this: Calculate the molecular mass of trichloromethane, a lso k n ow n a s Fre o n - 11, w h o s e c o n d e n s e d structural formula is CCl3F. Until recently, it was used as refrigerant. The structural of a molecular Freon-11 is shown below: The answer must be 136 g/mole  gram/g Determining Mass Gram (g) = mole x ------------ mole Mass (element) = mole (element) x atomic mass (element) Mass (ionic compound) = mole (ionic compound) x formula mass Mass (molecular compound) = mole (molecular compound) x molecular mass  gram (g) Determining Mole Mole = ----------------- gram (g)/mole mass (element) Mole (element) = -------------------------- atomic mass (element) mass (ionic compound) Mole (ionic compound) = ----------------------------- formula mass mass (molecular compound) Mole (molecular compound) = --------------------------------- molecular mass Determining Formula mass or Molecular mass Gram (g)/mole = -------- gram mole mass (element) atomic mass (element) = --------------------- mole (element) mass (ionic compound) Formula mass = ----------------------------- mole (ionic compound) mass (molecular compound) Molecular mass = --------------------------------- mole (molecular compound) Determining Molecular and Empirical Formula When a new chemical compound, such as a potential new pharmaceutical is synthesized in the laboratory or isolated from a natural source, chemists determine its elemental composition, its empirical formula, and its structure to understand its properties. In this lesson, the focus on how to determine the empirical formula of a compound and then use it to determine the molecular or formula mass of the compound is known. Example: What are the percentages of carbon, hydrogen and oxygen in sucrose, molecular formula C12H22O11. From the molecular formula (C12H22O11) the mass of carbon, hydrogen and oxygen can be computed. To compute for the: mass = mole x atomic mass Mass of carbon = moles of carbon x atomic mass of carbon = 12 moles x 12 g/mole = 144 grams Mass of hydrogen = moles of hydrogen x atomic mass of hydrogen = 22 moles x 1 g/mole = 22 grams Mass of oxygen = moles of oxygen x atomic mass of oxygen = 11 moles x 16 g/mole = 176 grams  Element Analysis by Computation Percent mass (gram) (%) Carbon 144 144 g/342 g x 100 = 42.1 Hydrogen 22 22 g/342 g x 100 = 6.4 Oxygen 176 176 g/342 g x 100 = 51.5 Total 342 100.0 Thus, the percent of carbon in sucrose (C12H22O11) is 42.1 %, the percent of hydrogen is 6.4 %, and the percent of oxygen is 51.5 %. Supposing the percentage of the elements in the unknown compound is given. Determine the empirical formula of sample problem: Sample Problem1: An unknown tablet subject for analysis and it was found out that it contains 40.92 % C, 4.58 % H, and 54.50 % O, by mass. The experiment ally determined molecular mass is 176 g/mole. What is the empirical and chemical formula for unknown tablet? Solution: Consider an arbitrary amount of 100 grams of the unknown tablet, so it will be: 40.92 grams C 4.58 grams H 54.50 grams O This would give us how many moles of each element? moles = mass/atomic mass Moles of carbon = 40.92 g/(12 g/mole) = 3.41 moles/3.40 moles` =1 x 3 = 3.0 = 3 Moles of hydrogen =4.58 g/(1 g/mole) = 4.58 moles/3.40 moles` =1.3 x 3 = 3.9 = 4 Moles of oxygen =54.50 g/(16 g/mole) = 3.40 moles/3.40 moles =1 x 3 = 3.0 = 3 The relative molar amounts of carbon and oxygen appear to be equal, but the relative molar amount of hydrogen is higher. Since atom cannot have "fractional" value in a compound, the need to normalize the relative amount of hydrogen to be equal to an integer (any of the natural numbers, the negative of these numbers, or zero). 1.333 would appear to be 1 and 1/3, so if we multiply the relative amounts of each atom by '3', we should be able to get integer values for each atom. Thus the empirical formula is C3H4O3 What about the chemical formula? The molecular mass was determined experimentally, it is 176 g/mole. What is the empirical mass of the empirical formula? The empirical mass is: 3 x atomic mass of carbon = 3 x 12 g/mole = 36 g/mole 4 x atomic mass of hydrogen = 4 x 1 g/mole = 4 g/mole 3 x atomic mass of oxygen = 3 x 16 g/mole = 48 g/mole 88 g/mole The molecular mass from the empirical mass is significantly lower than the experimentally determined value. What is the ratio between the two values? (176 g/mole/88 g/mole) = 2.0 Thus, it would appear that our empirical formula is essentially one half the mass of the actual molecular mass. If we multiplied the empirical formula by '2', then the molecular mass would be correct. Thus, the moleculaf formula is: 2(C3H4O3) = C6H8O6. This is the formula of ascorbic acid therefore the unknown table is ASCORBIC ACID. Sample Problem 2. A laboratory technician ran an analysis on one of the germicides stocked by a janitorial services. The technician determined that the effective ingredient in the germecide was a compound that was 33% sodium, 36% arsenic, and 31% oxygen. What is the formula and name of this compound? Solution Solve first for the moles of each elements: Na: 33 grams/23 grams/mole = 1.5 moles/0.5 mole = 3 As: 36 grams/75 grams/mole = 0.5 mole/0.5 mole = 1 O: 31 grams/16 grams/mole = 2.0 moles/0.5 mole = 4 The formula of the compound is: Na3AsO4 (sodium arsenate) Sample Problem 3. The major air pollutant of coal-burning power plants is a colorless, pungent gaseous compound containing only sulfur and oxygen. Chemical analysis of a 1.078 g sample of this gas showed that it contained 0.540 g of S and 0.538 g of O.What is the empirical formula of this compound? Solution 0.540 g Moles S = ---------- = 0.0169 mole/0.0169 mole = 1 32 g/mole 0.538 g Moles O = ---------- = 0.0336 mole /0.0169 mole = 2 16 g/mole Therefore the empirical formula of the compound is SO2 (sulfur dioxide) Sample Problem 4. Benzene has the empirical formula CH. Its molecular mass is 78g/mole. What is its molecular formula? Solution The empirical mass of CH C: 1 x 12 g/mole = 12 g/mole H: 1 x 1 g/mole = 1 g/mole 13 g/mole Ratio of molecular mass and empirical mass = 78 g/mole/13 g/mole = 6 Therefore, the molecular formula of benzene is C6H6. From the given in table below differentiate empirical formula from molecular formula Answer: The empirical formula Empirical Molecular gives the smallest whole number Formula Formula ra t i o b et we e n e l e m e n t s i n a NO2 N2O4 compound while the molecular CH C6H6 formula gives the actual whole CH3O C2H6O2 number ratio between elements in a compound. The law of definite proportions states that: “a chemical compound always contains the same proportion of elements by mass; that is, the percent composition—the percentage of each element present in a pure substance—is constant.” Note: although there are exceptions to this law. Solve this problems. 1. What are the percentages of each elements in calcium permanganate ? 2. Herbadox® is a selective pre-emergent herbicide which contains 330 g/L Pendimethalin (C 13 H 19 N 3 O 4 ). With its contact action, it effectively inhibits the root and shoot growth of grasses such as Eleucine indica, Rottboellia exaltata, Echinochloa sp., Ischaemum rugosum. Compute for the percentages of each elements in the formula. 3. The chemist ran an analysis found out that a certain unknown chemical compound contain 14.39% calcium, 39.57% manganese, and 46.04% oxygen. What is the formula and name of this unknown chemical compound? Solve this problems. 4. In areas where temperature get extremely cold, people must take special precautions to make sure machinery runs properly. One compound containing 83% rubidium, 16% oxygen, and 1% hydrogen is used in storage batteries designed for use in low temperatures. What is the empirical formula for this compound? 5. What is the empirical formula of a compound if a sample contains 0.130 g of nitrogen and 0.370 g of oxygen? 6. Cyclobutane has the empirical formula CH2. Its molar molar mass is 42g. What is its molecular formula?