Determining Chemical Formulas
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1 Deterining Cheical Forulas Joseph Proust proposed the law of constant coposition. This law states that a copound contains eleents in certain fixed proportions (ratios) and in no other cobinations, regardless of how the copound is prepared or where it is found in nature. Ex: one olecule of ethane, CH 4(g), always contains one ato of carbon and four atos of oxygen. Percentage Coposition is the percentage, by ass, of each eleent in a copound. %H H 100% sa p le C %C 100% sa p le Epirical forula a forula that gives the lowest ratio of atos in a copound Deterining the Epirical Forula Ex: The percentage coposition of a copound is 69.9% iron and 30.1% oxygen. What is the epirical forula of the copound? Step 1: List Given Values Fe = 69.9% O = 30.1% Step 2: Calculate Mass () of Each Eleent in 100-g Saple Fe 100 g 100 Fe = 69.9 g 301. O 100 g 100 O = 30.1 g Step 3: Convert Mass () into Aount (n) 1 ol Fe nfe 69.9 g Fe g Fe n Fe = 1.25 ol Fe
2 n = 30.1 g O O n O = 1.88 ol O 1 ol O g O Step 4: State Aount Ratio n Fe : n O = 1.25 : 1.88 Step 5: Calculate Lowest Whole-Nuber Aount Ratio Divide all the aounts by the sallest aount. Here the sallest aount is 1.25 ol Fe. n Fe :n O = 1.00 : 1.50 Notice that this ratio has one nuber that is not a whole nuber. To obtain a whole nuber, ultiply both nubers by 2. n Fe :n O = 2 : 3 Therefore, the epirical forula of the copound is Fe 2 O 3. Ex 2: The percentage coposition of a copound is 21.6% sodiu, 33.3% chlorine, and 45.1% oxygen. What is the epirical forula of the copound?
3 Molecular forula a forula that indicates the actual nubers of atos in one olecule of a copound. Deterining the Molecular Forula The epirical forula of a copound is CH 3 O, and its olar ass is 93.12g/ol. What is the olecular forula of the copound? Step 1: List given values Epirical forula of copound = CH 3 O M copound = g/ol Step 2: Deterine Molar Mass of Epirical Forula MCH3O = 1(12.01 g/ol) + 3(1.01 g/ol) + 1(16.00 g/ol) MCH3O = g/ol Step 3: Deterine Ratio of Molar Mass of Copound to Molar Mass of Epirical Forula The Mc o p o und Mepirical foula ratio gives the factor by which the olar ass of the copound is greater than the olar ass of the epirical forula. Mc o p o und g/ol M C H 3O g/ol M M c o p o und C H 3O 3 Step 4: Calculate Molecular Forula Use the ratio deterined in step 3 to calculate the olecular forula. The olar ass of the copound is three ties greater than the olar ass of its epirical forula. Multiply the subscripts of the epirical forula by 3 to obtain the subscripts of the olecular forula. olecular forula = 3(epirical forula) = 3(CH 3 O) olecular forula = C 3 H 9 O 3 The olecular forula of the copound is C 3 H 9 O 3.
4 Ex2: The percentage coposition is 40.03% carbon, 6.67% hydrogen and 53.30% oxygen. The olar ass of the copound is found to be g/ol. What is the olecular forula of the copound?
5 Calculating Percentage Coposition by Mass Calculate the percentage coposition of carbon dioxide, CO 2(g). Step 1: Calculate Total Mass of Each Eleent in Copound C = u O = 2(16.00 u) O = u Step 2: Calculate Molecular Mass (or Foula Unit Mass) of Copound CO2 = C + O = u u CO2 = u Step 3: Calculate Percentage Coposition by Mass of Copound % C = C CO2 % C = u u % C = 27.29% % O = O CO2 % O = u u % O = 72.71% The percentage coposition by ass of carbon dioxide is 27.29% carbon and 72.71% oxygen. Ex2: Calculate the percentage coposition by ass of potassiu sulfate, K 2 SO 4(s).
6 Worksheet 2.3: Deterining Cheical Forulas 1. Calculate the epirical forula of a copound that is found to contain 2.2% hydrogen, 26.7% carbon, and 71.1% oxygen. 2. The percentage coposition of a copound is 35.9% aluinu and 64.1% sulfur. What is the epirical forula of the copound? 3. A fat that is used to ake soap contains 76.5% carbon, 12.2% hydrogen, and 11.3% oxygen by ass. Deterine the olecular forula of the fat if its olar ass is g/ol. 4. The percentage coposition of nicotine is 74.0% carbon, 8.7% hydrogen, and 17.3% nitrogen. Its olar ass is g/ol. What is the olecular forula of nicotine? 5. Calculate the percentage coposition by ass of each of the following copounds: a. C 6 H 8 O 6(s) b. Al 2 O 3(s) c. Zn(NO 3 ) 2(s)
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Micro World atoms & molecules Laboratory scale measurements Atomic mass is the mass of an atom in atomic mass units (amu) By definition: 1 atom 12 C weighs 12 amu On this scale 1 H = 1.008 amu 16 O = 16.00
2 The Structure of Atoms
CHAPTER 4 2 The Structure of Atoms SECTION Atoms KEY IDEAS As you read this section, keep these questions in mind: What do atoms of the same element have in common? What are isotopes? How is an element
EXPERIMENT 12: Empirical Formula of a Compound
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AS1 MOLES. oxygen molecules have the formula O 2 the relative mass will be 2 x 16 = 32 so the molar mass will be 32g mol -1
Moles 1 MOLES The mole the standard unit of amount of a substance the number of particles in a mole is known as Avogadro s constant (L) Avogadro s constant has a value of 6.023 x 10 23 mol -1. Example
Stoichiometry Review
Stoichiometry Review There are 20 problems in this review set. Answers, including problem set-up, can be found in the second half of this document. 1. N 2 (g) + 3H 2 (g) --------> 2NH 3 (g) a. nitrogen
CHEM 120 Online: Chapter 6 Sample problems Date: 2. Which of the following compounds has the largest formula mass? A) H2O B) NH3 C) CO D) BeH2
CHEM 120 Online: Chapter 6 Sample problems Date: 1. To determine the formula mass of a compound you should A) add up the atomic masses of all the atoms present. B) add up the atomic masses of all the atoms