What is the molecular formula of a material for which complete combustion of a unknown mass gives #3.38*g# of #CO_2(g)#, and #0.69*g# of water. An #11.6*g# mass of this gas gives a volume of #10.0*L# at #0# #""^@C# of #10*L#?

Answer 1

You probably have......#"ethylene"#

The ideal gas equation is the only foundation upon which we can examine the molecular formula. The combustion data you provided were not kosher; neither the mass of the combusted gas nor the water content of the combustions were disclosed.

If #PV=nRT#, then #1*atmxx10L=(11.6*g)/"Molar mass"xx0.0821(L*atm)/(K*mol)xx298*K#

OR

#"Molar mass"=(11.6*g)/(1*atmxx10*L)xxxx0.0821(L*atm)/(K*mol)xx298*K=28.4*g*mol^-1#.
Now this molecular mass is consistent with that of #"ethylene"#, #H_2C=CH_2# whose molecular mass is #28.0*g*mol^-1#.

Nevertheless, I don't believe you have addressed all the relevant information in this query.

Had we calculated as molecular mass of #26*g*mol^-1#, or #30*g*mol^-1#, what would be the likely identity of the gas under the given scenario?
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Answer 2

The molecular formula is #"C"_2"H"_2#.

This problem relates to the empirical formula/molecular formula.

(a) Calculate the moles of #"C"#.
#"Moles of C" = 3.38 color(red)(cancel(color(black)("g CO"_2))) × ("1 mol CO"_2)/(44.01 color(red)(cancel(color(black)("g CO"_2)))) × "12.01 g C"/(1 "mol C") = "0.076 80 mol CO"_2#
(b) Calculate the moles of #"H"#.

I take it that you intended 0.69 grams of water.

#"Moles of H" = 0.69 color(red)(cancel(color(black)("g H"_2"O"))) × (1 color(red)(cancel(color(black)("mol H"_2"O"))))/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) × "2 mol H"/(1 color(red)(cancel(color(black)("mol H"_2"O")))) = "0.0766 mol H"#

Determine the molar ratios (c).

I like to compile the calculations into a table from this point on.

#bb("Element"color(white)(Xll) "Moles"color(white)(m) "Ratio" color(white)(m)"Integers")# #color(white)(mm)"C" color(white)(mmmll)0.07680color(white)(X)1.003color(white)(mmml)1# #color(white)(mm)"H" color(white)(mmmll)0.0766 color(white)(mll)1 color(white)(Xmmmml)1#
The empirical formula is #"CH"#.

(d) Determine the gas's molar mass.

Since you don't specify the gas's temperature or pressure, I'll assume STP (1 bar and 0 °C).

#pV = nRT = m/MRT#
#M = (mRT)/(pV) = ("11.6 g" × "0.083 14" color(red)(cancel(color(black)("bar·L·K"^"-1")))"mol"^"-1" × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar"))) × 10 color(red)(cancel(color(black)("L")))) = "26.3 g/mol"#

(d) Determine the gas's formula.

The empirical formula mass of #"CH"# is 13.02 u.

The gas has a molecular mass of 26.3 u.

An integral multiple of the empirical formula mass is required for the molecular mass.

#"MM"/"EFM" = (26.3 color(red)(cancel(color(black)("u"))))/(13.02 color(red)(cancel(color(black)("u")))) = 2.02 ≈ 2#

The empirical formula must be doubled in the molecular formula.

#"MF" = ("EF")_2 = ("CH")_2 = "C"_2"H"_2#
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Answer 3

The molecular formula of the material can be determined using the given information and stoichiometry.

  1. Find the moles of CO2 and H2O produced:

    • Moles of CO2 = 3.38 g / 44.01 g/mol
    • Moles of H2O = 0.69 g / 18.015 g/mol
  2. Determine the limiting reactant:

    • Calculate the moles of C and H in CO2 and H2O.
    • The smaller value is the limiting reactant.
  3. Calculate the moles of the unknown material:

    • Use the moles of the limiting reactant and the stoichiometry of the reaction.
  4. Find the molecular formula:

    • Divide the molar mass of the unknown material by the molar mass calculated in step 3 to find the molecular formula ratio.
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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