Example: New empirical formula of your own compound sugar (C

O = \(\frac \) ? Mass = \(\frac \) ? Molecule wt

Empirical formula The empirical formula of a compound may be defined as the formula which gives the simplest whole number ratio of atoms of the various elements present in the molecule of the compound. _sixH_{a dozen}O₆), is CH₂O which shows that C, H, and O are present in the simplest ratio of 1 : 2 : 1. Rules for writing the empirical formula The empirical formula is determined by the following steps :

1. Separate the newest percentage of for each and every facets by the their atomic mass. This gives the brand new cousin number of moles of several points expose about compound.
2. Divide the fresh new quotients acquired regarding the significantly more than step from the tiniest of those so as to get a simple proportion regarding moles of numerous factors.
3. Multiply the new rates, so gotten of the a suitable integer, if necessary, in order to get entire count proportion.
4. In the end jot down brand new signs of the numerous points front by the side and place the above mentioned amounts just like the subscripts on down right hand corner of each icon. This will represent this new empirical formula of your compound.

Example: A compound, toward study, gave the second composition : Na = cuatro3.4%, C = 11.3%, O = forty five.3%. Determine its empirical formula [Atomic masses = Na = 23, C = several, O = 16] Solution:

O₃

_{Determination molecular formula : Molecular formula = Empirical formula ? n n = \(\frac \) Example 1: What is the simplest formula of the compound which has the following percentage composition : Carbon 80%, Hydrogen 20%, If the molecular mass is 30, calculate its molecular formula. Solution: Calculation of empirical formula :}

_{? Empirical formula is CH3. Calculation of molecular formula : Empirical formula mass = 12 ? 1 + 1 ? 3 = 15 n = \(\frac =\frac \) = 2 Molecular formula = Empirical formula ? 2 = CH3 ? 2 = C2H6.}

Example 2: On heating a sample of CaC, volume of CO₂ evolved at NTP is 112 cc. Calculate (i) Weight of CO₂ produced (ii) Weight of CaC taken (iii) Weight of CaO remaining Solution: (i) Mole of CO₂ produced \(\frac =\frac \) mole mass of CO₂ = \(\frac \times 44\) = 0.22 gm (ii) CaC > CaO + CO₂(1/200 mole) mole of CaC = \(\frac \) mole ? mass of CaC = \(\frac \times 100\) = 0.5 gm (iii) mole of CaO produced = \(\frac \) mole mass of CaO = \(\frac \times 56\) = 0.28 gm * Interesting by we can apply Conversation of mass or wt. of CaO = wt. of CaC taken – wt. of CO₂ produced = 0.5 – 0.22 = 0.28 gm

Example 3: If all iron present in 1.6 gm Fe₂ is converted in form of FeSO₄. (NH₄)₂SO₄.6H₂O after series of reaction. Calculate mass of product obtained. Solution: If all iron will be converted then no. of mole atoms of Fe in reactant product will be same. ? Mole of Fe₂ = \(\frac =\frac \) mole atoms of Fe = 2 ? \(\frac =\frac \) mole of FeSO₄. (NH₄)₂SO₄.6H₂O will be same as mole atoms of Fe because one atom of Fe is present in one molecule. ? Mole of FeSO₄.(NH₄)₂.SO₄.6H₂ = \(\frac \times 342\) = 7.84 gm.