Advertisements
Advertisements
प्रश्न
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.
उत्तर
Given mass of M: 11.2 grams
Atomic weight of M: 56 g/mol
Moles of M = `"Mass of M"/"Atomic weight of M"`
= `(11.2 " g")/(56 " g/mol")`
= 0.2 mol
Given mass of N: 4.8 grams
Atomic weight of N: 16 g/mol
Moles of N = `"Mass of N"/"Atomic weight of N"`
= `(4.8 " g")/(16 " g/mol")`
= 0.3 mol
or
M : N = 2 : 3
Empirical formula is M2N3
APPEARS IN
संबंधित प्रश्न
Identify the term or substance based on the descriptions given below:
The property by virtue of which the compound has the same molecular formula but different structural formulae.
Ethane burns in oxygen to form CO2 and H2O according to the equation:
`2C_2H_6+7O_2 -> 4CO_2 + 6H_2O`
If 1250 cc of oxygen is burnt with 300 cc of ethane.
Calculate:
1) the volume of `CO_2` formed
2) the volume of unused `O_2`
Explain the terms empirical formula and molecular formula.
Give example of compound whose:
Empirical formula is different from the molecular formula.
A compound of lead has following percentage composition, Pb = 90.66%, O = 9.34%. Calculate empirical formula of a compound. [Pb = 207, O = 16]
The compound A has the following percentage composition by mass: C =26.7%, O = 71.1%, H = 2.2%.
Determine the empirical formula of A.(Answer to one decimal place) (H=1,C=12,O=16)
Determine the empirical formula of the compound whose composition by mass is 42% nitrogen, 48% oxygen and 9% hydrogen.[N=14,O=16,H=1]
The percentage composition of sodium phosphate, as determined by analysis is : 42.1% Na, 18.9% P, 39% of O. Find the empirical formula of the compound.
[H =1, N =14, Na = 23, P = 31, Cl = 35.5, Pt = 195]
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.
Calculate:
Write the empirical formula of C8H18.
