weighed sample of chromium oxide, Cr,0, was analyzed by treating it with hydroge

Question

weighed sample of chromium oxide, Cr,0, was analyzed by treating it with hydrogen peroxide, which converted all the chromium oxide to chronmate ion, Cr042 2 Cr042-(aq) + 5H20 Cro, (s) + 3H202 (aq) + 40H. (aq) The solution containing the chromate ion from the sample was diluted to a volume of 0.500 L Beer's Law plot was made a standard C042-solution using cells with b = 1.00 cm was found to the slope given below, the intercept of this plot was negligible. Fill in the blanks below. Show your work in the space at the bottom of the page. Slope of the Beer's Law calibration plot Molar absorption coefficient (E) of CrO, Mass of the sample of Cr,O, used to make 0.500 L of solution Absorbance of solution made from your sample of Cr,O Molarity of Cro, in the solution made from your sample of Cr,O, Moles of CrO, present in solution made from your sample of Cr,O, Moles of Cr in solution made from your sample of Cr,O 267 L/mol 1.9400 g 12.965 Mass of Cr in solution made from your sample of Cr,O Experimental percentage by mass of Cr in your sample of Cr,O

Explanation / Answer

A = e*l*C

A = absorbance of sample

e is the molar absorptivity , the typical units are 1/M-cm

l size of cuvette, typically reported in cm

c is the molar concentration, in mol per ltier or M

get e

A = e*l*C

slope = e*l, if l = 1 cm, then

e = 267 L/mol-cm

Q2.

find [CrO4-2]

A = 267 * M

12.965 = 267 * M

M = 12.965 /267

M = 0.04855

now...

mol of CrO4-2 = M*V = (0.04855)(0.5) = 0.024275 mol of CrO4-2

mol of Cr in solution --> 1 mol of Cr per 1 mol of CrO4-2

mol of Cr = 0.024275 mol

mass of Cr = mol*MW = (0.024275)(51.9961) =1.2622 g of Cr

% mass of Cr = mass of Cr / Total mass * 100% = 1.2622 / 1.94*100 = 65% is Cr

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