Advertisements
Advertisements
प्रश्न
Find the empirical formula and the molecular formula of an organic compound from the data given below :
C = 75.92%, H = 6.32% and N = 17.76%
The vapour density of the compound is 39.5.
[C = 12, H = 1, N = 14]
Advertisements
उत्तर
| Element | % | |
| C | 75.92 | `75.92/12 = 6.32/1.26 = 5` |
| H | 6.32 | `6.32/1 = 6.32/1.26 = 5` |
| N | 17.76 | `17.76/14 = 1.26/1.26 = 1` |
Empirical formula ; - C5H5N
Molecular weight = 2 × Vapour Density
= 2 × 39.5
= 79
Empirical formula weight = C × 5 + H × 5 + N × 1
= 12 × 5 + 1 × 5 + 14 × 1
= 60 + 5 + 14
= 79
`n = "Molecular weight"/("Empirical" "formula" "weight" )`
∴ Molecular formula = Empirical formula = C5H5N
APPEARS IN
संबंधित प्रश्न
A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapor density is 29, find its molecular formula. [C = 12, H = 11]
Explain the terms empirical formula and molecular formula.
Give the empirical formula of C6H6.
Give the empirical formula of C2H2
Calculate the empirical formula of the compound having 37.6% sodium, 23.1% silicon and 39.3% oxygen.(Answer correct to two decimal places) (O = 16, Na = 23, Si = 28)
Calculate the percentage of water of crystallization in CuSO4. 5H2O
(H = 1, O = 16, S = 32, Cu = 64)
Give the empirical formula of C6H18O3
Silicon (Si = 28) forms a compound with chlorine (Cl = 35.5) in which 5.6 g of silicon combines with 21.3 g of chlorine. Calculate the empirical formula of the compound.
A compound has the following percentage composition by mass: carbon 14.4%, hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. Work correctly to 1 decimal place. (H = 1; \[\ce{C}\] = 12; \[\ce{Cl}\] = 35.5)
Pratik heated 11.2 grams of element ‘M’ (atomic weight 56) with 4.8 grams of element 'N' (atomic weight 16) to form a compound. Find the empirical formula of the compound obtained by Pratik.
