Making Formulas

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## Making Formulas

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**Making Formulas**% Composition, Empirical and Molecular Formulas**Percent Composition**Percent composition of a compound refers to the percentage by mass. • In other words, the percentage composition of water shows what % (by mass) is Hydrogen, and what % is Oxygen. Percentage composition by mass does not tell us the ratio by which the atoms combine. (Does not tell us the subscripts of chemical formulas)**% Composition Example**Jim goes to a school that has 258 total students. If 164 of the students are boys, what percentage of the student body is made up of girls?**% Example**In this example, you found the percent composition by dividing one part by the total and multiplying by 100. part % composition = -------------- x 100 whole % Girls = (258-164) x 100 258 % Girls = 36.4**Percent Composition**Finding the percentage composition by mass works the same way. • Divide the mass that one element contributes to the compound by the mass of the entire compound. mass from element % Composition = --------------------------- x 100total mass of the compound**% Composition Example**What percentage of the mass of carbon dioxide (CO2) is made up by the carbon? Mass of Carbon = 12.01 x 1 = 12.01 g Mass of Oxygen = 16.00 x 2 = 32.00 g part % comp. of Carbon = --------------- x 100 whole 12.01 g % comp. of Carbon = --------------- x 100 44.01 g Answer = 27.3%**% Composition Example**What is the percentage composition of carbon in glucose (C6H12O6) ?**Percent Composition and Formulas**• Percent composition only tells you what percentage of the compound that a particular atom is. It doesn’t tell you how many atoms are in that compound. • HOWEVER, it does help you to determine that compound’s empirical formula.**Empirical Formula**The simplest whole number ratio of atoms in a compound**Empirical Formulas**Identify each of the following as empirical or molecular: • CH4 • H2O • C2H6 • NaOH • C6H12O6**Empirical Formulas**Identify each of the following as empirical or molecular: • CH4 - Empirical • H2O - Empirical • C2H6 - Molecular • NaOH- Empirical • C6H12O6 - Molecular**Steps to Calculate Empirical Formulas**• Convert all given masses to moles (so they can be compared) • Find the lowest whole number ratio in moles by dividing by the smallest number of moles. • This whole # value becomes the subscript. The subscripts in the formula will reflect the smallest whole number ratio between the moles of each of the elements present.**Example**A sample of urea is decomposed into its elements. The following products are measured: Nitrogen: 1.121 g Hydrogen: 0.161 g Carbon: 0.480 g Oxygen: 0.640 g**1. Find the number of moles of each element present:**Nitrogen: 1.121 g x ( )= hydrogen: .161 g x ( )= carbon: .480 g x ( )= oxygen: .640 g x ( )= 0.08 mol N 0.16 mol H 0.04 mol C 0.04 mol O**0.08 mol N**0.16 mol H 0.04 mol C 0.04 mol O / 0.04 mol = / 0.04 mol = / 0.04 mol = / 0.04 mol = 2.0 N 4.0 H 1.0 C 1.0 O 2. Find the lowest whole number ratio in moles (the empirical formula)**2.0 N**4.0 H 1.0 C 1.0 O N2H4CO 3. The subscripts in the formula will reflect the smallest whole number ratio between the moles of each of the elements present.**Practice Problem**What is the empirical formula for a compound which contains 0.0134 g of iron, 0.00769 g of sulfur and 0.0115 g of oxygen?**Practice Problem**A compound is 92.24% Carbon and 7.76% Hydrogen. The molar mass of the molecular compound is 78.1 g. Calculate the empirical formula and molecular formula of the compound.**Mole Airlines Flight 1023**• Scenario: Read with your partner • Your Job: • Determine empirical formulas (Table 3) • Match your empirical formulas with the substances listed in Table 1 • Compare the completed Table 3 with Table 2 to complete the Victim Identification Form • Identify the murderer, victim, and explosive substance.