Calculate the mass of 1 a.m.u in grams.
In a reaction
Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2-5 mol B
(v) 2·5 mol A + 5 mol B
According to the given reaction:
(i) 1 atom of A will react with molecules of B2 = 1
300 atoms of A will react with molecules of B2 = 300
But molecules of B2 actually available = 200
∴ B2 is the limiting reactant.
(ii) 1 mol of A reacts with 1 mol of B
∴ 2 mol of A will react with 2 mol of B.
Hence A is the limiting reactant.
(iii) 100 atoms of A will react with 100 molecules of B. Hence there are no limiting reactants.
(iv) 2·5 mol of B will react with 2·5 mol of A. Hence B is the limiting reagent.
(v) 2·5 mol of A will react with 2·5 mol of B.
Hence A is the limiting reagent.
The following data were obtained when dinitrogen and dioxygen react together to form different compounds:
Mass of dinitrogen Mass of dioxygen
(i) 14g 16g
(ii) 14g 32g
(iii) 28g 32g
(iv) 28g 80g
which law of chemical combination is obeyed by the above experimental data? Give its statement?
A welding fuel gas contains carbon and hydrogen only. Burning a small space of it in oxygen gives 3.38 g carbon dioxide, 0·690g of water and no other products. A volume of 10·0L (measured at STP) of this welding gas is found to weigh 11·6g. Calculate: (i) empirical formula (ii) molar mass of the gas and (iii) molecular formula.
i.1mole(44g) of CO2 contains 12g of carbon.
3.38 g of CO2 will contain carbon = (12g/44g) x3.38g =0.9217g
18g of water contains 2g of hydrogen
Therefore 0.690g of water will contain hydrogen=2g/18gx 0.690 =0.0767 g
Since carbon and hydrogen are the only constituents of the compounds, the total mass of the compounds is:
Percent of C in the compound = (0.9217/.9984) x100 =92.32%
Percent of H in the compound =(0.0767/0.9984) x100=7.68%
Moles of carbon in the compound =92.32/12.00 =7.69
Moles of hydrogen in the compound= 7.68/1=7.68
Since, ration of carbon to hydrogen in the compound=7.69:7.68=1:1
Hence, the empirical formula of the gas is CH.
Weight of 10.0L of the gas (at S.T.P)=11.6 g
Therefore, weight of 22.4L of gas at STP =(11.6/10.0L) x22.4L
Hence, the molar mass of the gas is 26.0 g.
iii) Empirical formula mass of CH =12+1 =13g
n= molar mass of gas/empirical formula mass of gas
Therefore, molecular formula of gas= (CH)n
What do you mean by significant figures?
What do you know about significant figures? What are the rules for determing the number of significant figures?
Significant figures. The total number of digits (including the last one, though its value is doubtful) in the number used to express the physical quantity is called the number of significant figures. For example, the reading 24 · 84 cm having three certain digits (2, 4 and 8) and one doubtful digit (4) has four significant figures.
Rules for determining the number of significant figures:
1. Except zero, all the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 are significant. For example,
4·25 has three significant figures.
71·435 has five significant figures.
2. All digits are significant including zeros if zero appears in between non-zero digits, e.g.,
4·02 has three significant figures.
4·005 has four significant figures.
6·0042 has five significant figures.
3. Zeros to the left of the first non-zero digit in a number are not significant, they merely indicate the positions of the decimal point. e.g.,
0·428 has three significant figures.
0·054 has two significant figures.
0·0006 has one significant figure.
4. Zeros to the right of the decimal point are significant. e.g.,
42·0 has three significant figures.
42·00 has four significant figures.
42·000 has five significant figures.
5. In exponential notation, the numerical portion gives the number of significant figures. e.g.,
1·58 × 10–1 has three significant figures.
4·243 × 105 has four significant figures.
3·24 × 107 has three significant figures.