Presentation on theme: "1 Chapter 7- Molecular Formulas. 2 Warm Up: Determine the name and molar mass of the following compounds. NaOH MnO 2 H 2 SO 4."— Presentation transcript:
1 Chapter 7- Molecular Formulas
2 Warm Up: Determine the name and molar mass of the following compounds. NaOH MnO 2 H 2 SO 4
3 How did you do? NaOH g/mol, sodium hydroxide MnO g/mol, Manganese (IV) oxide H 2 SO g/mol, Sulfuric acid Be sure you go back and review the basics…
4 Molecular Formula Activity This activity was designed to help you remember some of the content we learned last semester. You will also be learning about formulas (empirical and molecular) and percent composition.
5 The Activity 1. Work in groups of 3 or Be sure that every word of the activity is read out loud…take turns. 3. Every person needs to fill in their sheet. 4. The bags containing LEGO molecules are on my desk, come get one when you are ready. Balances are out.
Warm Up- Write the formulas 1. Calcium hydroxide 2. Carbon monoxide 3. Iron (III) chloride 4. Copper (II) nitrate 5. Sulfur tetroxide 6
Finish up the activity…. Questions so far?? 7
8 What is Percent Composition? Two ways of describing the composition of a compound. HCl 1. How many atoms of each element 2. The mass of each element in the compound (percentage).
9 How do I calculate Percent Composition? 1. Calculate the molar mass 2. Mass of element x 100= % composition Total Molar Mass 3. Add all the %compositions to make sure they total 100%.
10 Lets find the percent composition of each element in water. H 2 O
11 Find the percent composition of each element in these compounds Li 3 N BaSO 3
12 Li 3 N 1. Molar mass Li3moles x 6.94 g/mol = 20.82g N1mol x g/mol = g 20.82g g = 34.83g
13 Mass percent Li 20.82g x 100 = 59.78% 34.83g Mass percent N 14.01g x 100 = 40.22% 34.83g 59.78% % = 100%
14 BaSO 3 1. Molar mass Ba1mol x 137.3g/mol = 137.3g S1mol x 32.07g/mol = 32.07g O3mol x 16.00g/mol = 48.00g Total =217.37g
15 BaSO 3 Mass percent Ba 137.3g x 100 = 63.16% g Mass percent S 32.07g x 100 = 14.75% g Mass percent O 48.00g x 100 = 22.08% g 63.16% % % =99.99% Close enough!!
16 K 2 Cr 2 O 7 1. Molar mass K2mol x 39.10g/mol = 78.20g Cr2mol x 52.0g/mol = g O7mol x 16.00g/mol = g Total =294.2g
17 K 2 Cr 2 O 7 Mass percent K 78.20g x 100 = 26.58% 294.2g Mass percent Cr g x 100 = 35.35% 294.2g Mass percent O 112g x 100 = 38.07% 294.2g 26.58% % % =100%
18 Look at the Percent Comp WS You can do these no problem right? Look at the problems at the bottom!
19 Once you determine the Percent composition you can do various calculations How much potassium can be obtained in the decomposition of 50.0 g of K 2 Cr 2 O 7 ?
20 How did we determine the chemical formula of your LEGO compound ? We take a known amount of the sample and take it apart. All the pieces are collected and weighed. This provides the mass of each type of element allowing us to determine the mass percent of each element. We use this information to determine the formula.
21 Example Suppose we have a compound that contains the elements C, H and O. Total mass = g. C = g H = g O = g
22 Convert grams to moles. (0.0806g C) x 1mol C = mol C 12.01g ( g H) x 1mol H = mol H 1.008g H (0.1074g O) x 1mol O = mol O 16.00g O
23 Now, look for the smallest number and divide! mol C mol H mol O This usually results in whole numbers- that tell us the ratio of atoms in the compound
24 Warm Up- Jan 11, 2013 Determine the percent composition of nitrogen in NO 3.
25 How did you do on % Comp HW? 1. K=24.74% Mn=34.76% O = 40.50% 3. Mg = 16.39% N = 18.89% O = 64.72% 5. Al=15.77% S= 28.11% O= 56.12% g O g Fe g Ag Questions??
26 Empirical Formula Symbols of the elements with subscripts showing the smallest whole-number mole ratio of the different atoms in the compound. CH 2 O 1:2:1 Mole ratio
27 Empirical vs. Molecular formulas Empirical formula- smallest whole-number ratio of the atoms in a compound. Molecular formula- gives the actual number of atoms in the compound.
28 What would be the empirical formulas? CH 2 O C 2 H 4 O 2 C 3 H 6 O 3 C 4 H 8 O 4
29 Example: Step 4 An oxide of aluminum is formed by the reaction of 4.151g Al with 3.692g of Oxygen. Determine the empirical formula of the compound.
30 K: Aluminum Oxide is formed- Al and O g of Al 3.692g of O UK: Empirical formula Al O
31 2. Determine the number of moles of each element g Al x 1mol = mol Al 26.98g Al 3.692g O x 1mol = mol O 16.00g O
32 3. Divide the number of moles of each element b the smallest number of moles mol Al= 1 mol Al mol O = 1.5 mol O Bummer! Not whole numbers! You cant have a half a mol! What can I multiply both numbers by and get whole numbers?
= 1 mol Al x 2 = = 1.5 mol O x 2 = Empirical formula- Al 2 O 3
34 Warm Up If the empirical formula is CO, write 3 molecular formulas that would be correct.
Determining Molecular Formulas If you know the empirical formula… CH 2 And are given the molar mass of the molecular formula… 42.09g/mol We want to figure out how many of the empirical formulas fit into the molecular formula. 35
You try… A compound has a molar mass of g/mol. The compound has an empirical formula of NO. Determine the molecular formula. 36
Sometimes you need to figure out the empirical formula first! A sample contains a compound that is 30.45% nitrogen and 69.55% Oxygen. The molar mass of the compound is 92.02g/mol. What is the molecular formula. a. Assume 100g- makes % into g b. Convert grams to moles c. Divide by the smallest # moles d. Write the Empirical formula and determine the molar mass. e. Divide the molar mass of the compound by the molar mass of the empirical formula. 37
Work on the WS… Homework is the formula review… Major quiz Wed/Thurs… 38