Complete combustion of 1.11 0 g of a gaseous hydrocarbon yields 3.613 g of carbon dioxide and one 1.109 g of water. 80.288 g sample of the hydrocarbon occupies a volume of 131 milliliters at 24.8°c and 753 mm hg. what is the molecular formula of the hydrocarbon?

 C4H6

Explanation:

we know that

n = P×V / RT

= (753 ) x (0.131 ) / ((62.36367) x (24.8 + 273.15) ) = 0.005309

molecular weight = (0.288) / (0.005309) = 54.25 g/mol

molecular weight  of CO_2 in terms of C

(3.613 ) / (44.00964 ) x (12.01078) = 0.986033g C

Molecular weight of H_2O in terms of H

(1.109 ) / (18.01532 ) x (2  / 1) x (1.007947) = 0.124096g H

Number of atoms of C

(0.986033) / (1.110) x (54.25) / (12.01078) = 4.012

(0.124096) / (1.110) x (54.25 ) / (1.007947) = 6.017

Round to the nearest whole numbers to find the molecular formula:  C4H6

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.