1) A 1.4326 g of sample of a new compound has been analyzed and found to contain

Question

1) A 1.4326 g of sample of a new compound has been analyzed and found to contain the following masses of elements, carbon, 0.7263 g; hydrogen, 0.08127 g; nitrogen, 0.1412 g; oxygen, 0.4838 g. Calculate the empirical formula of the compound.

2) Using average atomic masses, calculate the number of atoms present in each of the following samples.

(a) 2.68 g of gold

(b) 0.000265 mol of platinum
(c) 0.000265 g of platinum
(d) 5.8 lb of magnesium
(e) 1.50 mL of liquid mercury (density = 13.6 g/mL)
(f) 4.45 mol of tungsten
(g) 4.45 g of tungsten

3) Calculate the number of moles of the indicated substance present in each of the following samples.

(a) 20.4 mg of nitrogen dioxide

(b) 1.76 g of copper(II) nitrate

(c) 2.47 g of carbon disulfide
(d) 4.92 g of aluminum sulfate
(e) 2.82 g of lead(II) chloride
(f) 61.3 g of calcium carbonate

Explanation / Answer

3 Ans:-

(a)First of all you have to turn mg into g, so you divide 21.4 by 1000 ( 1000 mg is equal to 1 g you do the conversion) so you have 0.0214 g.

Then you have to find de Molecular Mass of NO2(Nitrogen dioxide)

(The values for nitrogen and oxygen can be found by looking at a periodic table. )
*N=14 g/mol
*O2=16x2=32 g/mol ( since O has a subscript of two multiply 16 by 2. )
*MM(molecular mass)= N+O2= 14 +32=46 g/mol
Then with the formula n=mass/MM
n=0.0214g/46 g/mol = 4.65 x 10 ^ -4

Thrid use the amount in grams we started out with, which is .0214g, and
divide that by the molar mass found which is 46.01g.

Your answer should be .0004651 moles.

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