O = \(\frac < 1> < 50>\) ? Mass = \(\frac < 1> < 50>\) ? 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_severalO₆), 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. Split the fresh part of for each and every facets by their nuclear bulk. This gives the newest relative amount of moles of various issue present regarding compound.
2. Separate this new quotients received about more than action of the smallest of them to get a straightforward ratio regarding moles of numerous issue.
3. Proliferate the new data, therefore received by the ideal integer, if necessary, to see entire count ratio.
4. Finally jot down this new symbols of the numerous facets front side from the front and place the above mentioned number since subscripts with the lower right-hand part of each icon. This will depict brand new empirical formula of one’s compound.

Example: A material, towards the data, gave another composition : Na = cuatrostep three.4%, C = 11.3%, O = 45.3%. Assess the empirical formula [Nuclear masses = Na = 23, C = 12, O = 16] Solution:

O₃

_{Determination molecular formula : Molecular formula = Empirical formula ? n n = \(\frac < Molecular\quad> < Empirical\quad>\) 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 = \(\frac < Molecular\quad> < Empirical\quad>=\frac < 30> < 15>\) = 2 Molecular formula = Empirical formula ? 2 = CH3 ? 2 = C2H6.}

Example 2: On heating a sample of CaC, volume of CO₂ evolved at NTP is latin dating sites uk 112 cc. Calculate (i) Weight of CO₂ produced (ii) Weight of CaC taken (iii) Weight of CaO remaining Solution: (i) Mole of CO₂ produced \(\frac < 112> < 22400>=\frac < 1> < 200>\) mole mass of CO₂ = \(\frac < 1> < 200>\times 44\) = 0.22 gm (ii) CaC > CaO + CO₂(1/200 mole) mole of CaC = \(\frac < 1> < 200>\) mole ? mass of CaC = \(\frac < 1> < 200>\times 100\) = 0.5 gm (iii) mole of CaO produced = \(\frac < 1> < 200>\) mole mass of CaO = \(\frac < 1> < 200>\times 56\) = 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 = \(\frac < 1.6> < 160>=\frac < 1> < 100>\) mole atoms of Fe = 2 ? \(\frac < 1> < 100>=\frac < 1> < 50>\) 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 = \(\frac < 1> < 50>\times 342\) = 7.84 gm.