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1.1.5 Types of Formula

1.1.5 Types of Formula. Explain the terms empirical formula and molecular formula Calculate empirical and molecular formula. You will be able to…. Empirical Formula. The empirical formula of a compound is the simplest formula which represents its composition

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1.1.5 Types of Formula

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  1. 1.1.5 Types of Formula • Explain the terms empirical formula and molecular formula • Calculate empirical and molecular formula You will be able to…

  2. Empirical Formula • The empirical formula of a compound is the simplest formula which represents its composition • It shows the elements present and the ratio of the amounts of the elements present • To find the empirical formula you need to find the ratio of the amounts of the elements present

  3. Worked example 1 Analysis showed that 0.6075 g of magnesium combines with 3.995 g of bromine to form a compound. [Ar: Mg, 24.3; Br, 79.9.] Find the molar ratio of atoms: Mg : Br 0.6075 : 3.995 24.3 : 79.9 0.025 : 0.050 Divide by smallest number (0.025): 1 : 2 Empirical formula is: MgBr2

  4. Worked Example 2 Analysis of a compound showed the following percentage composition by mass: Na: 74.19%; O: 25.81%. [Ar: Na, 23.0; O,16.0.] 100.0 g of the compound contains 74.19 g of Na and 25.81 g of O. Find the molar ratio of atoms: Na : O 74.19 : 25.81 23.0 : 16.0 3.226 : 1.613 Divide by smallest number (1.613): 2 : 1 Empirical formula is: Na2O

  5. Questions 1. An 18.000g sample of magnesium oxide was found by reduction to contain 10.800g of magnesium. Calculate the empirical formula 2. A 9.435g sample of a chloride was found to contain 3.400g of calcium. Calculate the empirical formula 3. A sample of a compound contains 91.87g Fe, 79.11g S and 157.90g O. Determine the empirical formula

  6. Molecular Formula • The molecular formula is a simple multiple of the empirical formula e.g • If the empirical formula is CH2O then the molecular formula could be CH2O, C2H4O2, C3H6O3 …….. CnH2nOn n can be calculated by knowing the empirical formula mass and the molar mass n = Molar mass Empirical Mass

  7. Worked Example A compound has an empirical formula of CH2 and a relative molecular mass, Mr, of 56.0. What is its molecular formula? Answer • empirical formula mass of CH2: = 12.0 + (1.0 × 2) = 14.0 • number of CH2 units in a molecule: = 56.0 = 4 14.0 • molecular formula: (4 × CH2) = C4H8

  8. Questions • Benzene has the empirical formula CH and a molar mass of 78. What is the molecular formula? • Which of the following is an empirical formula and which a molecular formula? C3H7 C3H6 C4H10

  9. Calculation of percentage composistion From the formula of a compound and the relative atomic masses, the percentage of each element in the compound can be determined This is formally termed the percentage composition by mass

  10. Example - JFe • Calculate the percentage composition of nitroglycerine C3H5O9N3 C=12.01 H=1.01 O= 16.00 N=14.01

  11. 1.1.6 Moles and Gas Volumes • Calculate the amount of substance, in moles, using gas volumes You will be able to…

  12. Avogadro’s Law Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules i.e. 100cm3 of hydrogen at some temperature and pressure contains exactly the same number of molecules as 100cm3 of carbon dioxide

  13. One mole of different gases • Note the same volume of each gas has a different mass

  14. Example CH4(g) + 2O2(g) CO2(g) + 2H2O(l) • The equation says you need twice as many molecules of oxygen as you do of methane. According to Avogadro’s Law, this means you will need twice the volume of oxygen as of methane. • So, if you have to burn 1 litre of methane, you will need 2 litres of oxygen. • You should also produce 1 litre of carbon dioxide • Because water is a liquid, we don’t know how much of that we get

  15. Units of volume • Note: 1 litre = 1dm3 = 1000cm3 n = V/24(dm3)

  16. Problems • Take the molar volume to be 24.0dm3 at rtp. • Calculate the mass of 200cm3 of chlorine gas (Cl2) at rtp b) Calculate the volume occupied by 0.16g of oxygen (O2) at rtp.

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