Advertisements
Advertisements
Question
A compound of X and Y has the empirical formula XY2. Its vapor density is equal to its empirical formula weight. Determine its molecular formula.
Advertisements
Solution
Given:
Empirical formula = XY2
Vapor density = Empirical formula weight
Molecular formula =?
Molecular weight = n(Empirical formula weight)
= 2 x V.D.
n(Empirical formula weight) = 2 x V.D.
Since Vapour density Empirical formula weight
n = 2
Molecular formula = 2(Empirical formula)
= 2(XY2)
Molecular formula = X2Y4
APPEARS IN
RELATED QUESTIONS
A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapor density is 29, find its molecular formula. [C = 12, H = 11]
Give the empirical formula of C6H6.
Give the empirical formula of CH3COOH
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)
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 a compound containing 47.9‰ K, 5.5‰ beryllium and 46.6‰ fluorine by mass.
In a compound of magnesium (Mg = 24) and nitrogen (N = 14), 18 g of magnesium combines with 7g of nitrogen.
Deduce the simplest formula by answering the following questions:
- How many gram-atoms of magnesium are equal to 18g?
- How many gram-atoms of nitrogen are equal to 7g of nitrogen?
- Calculate the simple ratio of gram-atoms of magnesium to gram-atoms of nitrogen and hence the simplest formula of the compound formed.
A compound is formed by 24 g of X and 64 g of oxygen. If the atomic mass of X = 12 and O = 16, calculate the simplest 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.
