The empirical formula is #"C"_3"H"_8"#. Show
Since the percentages add up to 100, we can assume that we have a 100.0 g sample of the compound, and the percentages become grams. Determine the Moles of Each Element #81.82cancel"g C"xx(1"mol C")/(12.011cancel"g C")="6.812 mol C"# #18.18cancel"g H"xx(1"mol H")/(1.00794cancel"g H")="18.04 mol H"# Determine the Mole Ratios #"C":##(6.812)/(6.812)="1.000"# #"H":##(18.04)/(6.812)="2.648"# Since the mole ratio for H cannot be rounded to a whole number, we must multiply it times a factor that will result in a whole number. #"H":##2.648xx3=7.944~~8# We must multiply both mole ratios times #3#. #"C":##1.000xx3=3.000"# The empirical formula is #"C"_3"H"_8"#. Given the percentage composition of HC as #C->81.82% and H->18.18%# So the ratio of number if atoms of C and H in its molecule can will be #C:H=81.82/12:18.18/1# #C:H=6.82:18.18# So the Empirical Formula of hydrocarbon is #C_3H_8# As the mass of one litre of hydrocarbon is same as that of #CO_2# The molar mass of the HC will be same as that of #CO_2# i.e #44g/"mol"# Now let Molecular firmula of the HC be #(C_3H_8)_n# Using molar mass of C and H the molar mass of the HC from its molecular formula is So #44n=44=>n=1# Hence the molecular formula of HC is #C_3H_8# Home > Community > A compound contains 81.8% C and 18.2% H by mass. What is the empirical formula? C 81.8/12= 7 H 18.2/1=18.2 Take the smallest ratio and divide all through C 7/7=1 H 18.2/7=3 1:3 Empirical formular CH3 which is Methyl More C 81.8/12= 7 H 18.2/1=18.2 Take the smallest ratio and divide all through C 7/7=1 H 18.2/7=3 1:3 Empirical formular CH3 which is Methyl Number of carbon atoms : Number of hydrogen atoms = (mass percent of carbon/Atomic mass of carbon) : Mass percent of hydrogen/Atomic mass of hydrogen) = 81.8/12 : 18.2/1.008 = 6.8167 : 18.0556 = 1 : 2.6487 = 3 : 7.95 = 3 : 8 (nearest positive integer) So, the empirical formula of the compound is C3H8. Hope, this helps. More Number of carbon atoms : Number of hydrogen atoms = (mass percent of carbon/Atomic mass of carbon) : Mass percent of hydrogen/Atomic mass of hydrogen) = 81.8/12 : 18.2/1.008 = 6.8167 : 18.0556 = 1 : 2.6487 = 3 : 7.95 = 3 : 8 (nearest positive integer) So, the empirical formula of the compound is C3H8. Hope, this helps. Chem Week Reports Related Posts0 Upvotes · 3 Comments 0 Upvotes · 2 Comments 0 Upvotes · 7 Comments 0 Upvotes · 8 Comments 0 Upvotes · 3 Comments 0 Upvotes · 5 Comments 0 Upvotes · 2 Comments Copyright@Qingdao ECHEMI Digital Technology Co., Ltd. TOP What is the empirical formula for 81.8 carbon and 18.2 hydrogen?What is the empirical formula? So, the empirical formula of the compound is C3H8.
What is the empirical formula of a compound that contains 82% C and 18 H?1 Answer. The empirical formula is C3H8 .
What is the empirical formula of a compound that has a carbon to hydrogen?Empirical formula gives the smallest or simplest ratio of atoms present in a compound. In order to convert to the simplest ratio, divide the given ratio with 2. This mole ratio is used as subscripts in the empirical formula. Hence, the empirical formula of the compound is CH3 C H 3 .
What is the empirical formula of a compound that has 92.24% carbon and 7.76% hydrogen?Both Benzene (C6H6, molar mass = 78.12g/mol) and acetylene (C2H2, molar mass = 26.04g/mol) have the same percent composition (92.24 mass% carbon and 7.76% hydrogen) and the empirical formula, CH.
|