Empirical and molecular formula class 11
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The document discusses empirical and molecular formulas. An empirical formula shows the simplest whole number ratio of atoms in a compound, while a molecular formula shows the exact number of each atom. To calculate molecular formula from empirical formula: 1) make a table with elements, percentages, atomic masses, moles, and simplest ratios; 2) the empirical formula mass is the sum of the atoms' masses; 3) divide the molar mass by the empirical formula mass to get the common factor for the molecular formula. Two examples are given to calculate empirical and molecular formulas from percentage compositions and molar masses.
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Empirical and molecular formula class 11
- 2. What is empirical and molecular formula ? Basically , An empirical formula represents the simplest whole number ratio of various atoms present in a compound whereas the molecular formula shows the exact number of different types of atoms present in a molecule of a compound. We can calculate molecular formula by empirical formula by using the following steps :- 1) draw a table of 5 columns , including first column as element , then percentage composition , atomic mass , no of moles and then finally simplest ratio. After that we will get the empirical formula mass Calculate it and then divide by the molar mass given. We will get the molecular formula.
- 3. Questions Given the percentage composition as H- 4.07 , C-24.27% AND Cl-71.65% calculate the empirical and the molecular formula given the molar mass of the compound as 98.98g. Answer, element % composition Atomic mass No of moles Simplest ratio H 4.07 1 4.07 4.07/2.02=2 C 24.27 12 2.02 2.02/2.02=1 Cl 71.65 35.5 2.01 2.02/2.01=1
- 4. Therefore empirical formula becomes CH2Cl Now calculating the molar mass of CH2Cl we get 49.5 n= molar mass/ empirical formula mass = 98.96/49.5=2 Therefore the formula becomes 2(CH2Cl)=C2H4Cl2 Q.2) Given Na=14.3% , S=9.97% ,H=6.22% ,O=69.5% and all the H2 is present as water of crystallisation find the empirical and molecular formula. element % composition Atomic mass No of moles Simplest ratio Na 14.3 23 0.622 2 S 9.97 32 0.311 1 H 6.22 1 6.22 20 O 69.5 16 4.34 14
- 5. So after solving the compound will be Na2SH20O14 Now since all the water is present as wate rof crystallisation the formula becomes Na2SO4.10H20. Q.3) Find the molecular formula of the compound that contains 56.36g of oxygen and 54.6 grams of P. if the molar mass of the compound is said to be 189.5g/mol. Hint:- 1) Find the empirical formula. 2)divide by molar mass to find the common factor. 3) multiply to get the molecular formula