Advertisements
Advertisements
प्रश्न
Find the percentage of mass water in Epsom salt MgSO4.7H2O.
Advertisements
उत्तर
Formula of EPSOM salt = MgSO4.7H2O
molecular mass of MgSO4.7H2O = atomic mass of Mg + atomic mass of S + 4 × atomic mass of O + 7 { 2 × atomic mass of H + atomic mass of O }
= 24 + 32 + 4× 16 + 7{ 2 × 1 + 16 } g/mol
= (24 + 32 + 64+ 126 ) g/mol
= 246 g/mol
molecular mass of total water = 7 × ( 2× atomic mass of H + atomic mass of O)
= 7 × 18 = 126 g/mol
now ,
% mass of H2O in EPSOM salt = {total molar mass of H2O/molar mass of Epsom salts }× 100
= {126/246 } × 100
= 12600/246
= 51.21 %
APPEARS IN
संबंधित प्रश्न
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.
A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapor density is 29, find its molecular formula. [C = 12, H = 11]
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`
Give the empirical formula of C6H6.
Give the empirical formula of: C6H12O6
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.
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)
