2.00 grams of a straight-chained hydrocarbon contains 1.68grams of carbon, the rest hydrogen. 1. Calculate the empirical formula of this hydrocarbon.?

1 Answer
Jun 27, 2018

Answer:

The empirical formula is #"C"_4"H"_9#.

Explanation:

The empirical formula is the simplest whole number molar ratio of the elements in the compound.

We must convert the masses of #"C"# and #"H"# to moles and then find the ratio.

Step 1. Calculate the mass of #"H"#

#"Mass of H" = "mass of hydrocarbon - mass of C" = "2.00 g - 1.68 g = 0.32 g"#

Step 2. Calculate the moles of #"C"# and #"H"#

#"Moles of C" = 1.68 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)("g C")))) = "0.1399 mol C"#

#"Moles of H" = 0.32 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "0.317 mol H"#

From this point on, I like to summarize the calculations in a table.

#bbul("Element"color(white)(m) "Mass/g"color(white)(m) "Moles"color(white)(m) "Ratio"color(white)(m)×4color(white)(m)"Integers")#
#color(white)(mm)"C" color(white)(XXXm)1.68 color(white)(Xmll)0.1399 color(white)(Xm)1color(white)(mmml)4color(white)(mmmml)4#
#color(white)(mm)"H" color(white)(XXXm)0.32 color(white)(mmll)0.317 color(white)(Xml)2.27color(white)(mm)9.08 color(white)(mmm)9#

The molar ratio is #"C:H = 4:9"#.

The empirical formula is #"C"_4"H"_9#.

Here is a video that illustrates how to determine an empirical formula.