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Question

A 0.3288- sample of a hydrocarbon (a compound of carbon and hydrogen) was burned in air. The sample was completely converted to 1.0562 g of carbon dioxide and 0.5046 g of water. Find the emperical formula of the compound.







Please if you are not gonna help me do not send me any answers.

Answer 1

X + O2 --> CO2 + H2O

you got 1.0562g CO2 (MW = 44 g/mole) ==> moles = .024
and .5046 g H2O (MW = 18 g/mole) ==> moles = 0.028

H2O/CO2 = 0.028/0.024 = 1.168

This corresponds to rougly 7 CO2 and 7 H2O

1.168 (*2*3)= 7

x + O2 --> 7 CO2 + 7 H2O

2C7H14 + 21O2 --> 14CO2 + 14H2O

The empirical formula is C7H14

Answer 2

Lancenigo di Villorba (TV), Italy

PROPANE, e.g. C3H8.

This essay belongs to Gas-Test's Determinations.
As you said, the process leads the COMPLETE BURNING of a Pure Hydrocarbon's Amount.
Stoichiometric considerations are sketched in the following and fundamental chemical equation :

CxHy(g) + [x + (y / 4)] O2(g) ---> x CO2(g) + (y / 2) H2O(g)

You rescued a Carbon Dioxide's Amount as 1.0562 g like

(Carbon Dioxide's Moles) = 1.0562 / 44.00 = 0.024 mol

and you rescued a Water's Amount as 0.5046 g like

(Water's Moles) = 0.5046 / 18.00 = 0.028 mol

so these Molar's Amount undergoes the Ratio's Relation

nC : nH = nCO2 : (2 * nH2O) = 0.024 : (2 * 0.028) =
= 0.024 : 0.056 = 3 : 8

The overwritten ratio is CLEARLY related to C3H8, e.g. Propane.

I hope this helps you.

Answer 3

detrimental Erin, you should no longer have appeared very not ordinary. you opt for us that could actually help you consisting of your try. isn't that slightly unethical? As in dishonest? =========== follow up ============ in case you employ Joseph's recommendations, you opt for some success. there is not any indication that those are ions in answer, and in the adventure that they were, they does no longer form hi, on account that hi is a superior acid and is thoroughly dissociated in aqueous answer. What he wrote isn't what your instructor expects for the answer. imagine "diatomic". Your second one is even extra difficult. what's meant by technique of Iron(II)? Is it iron metallic? yet iron interior the metallic state has an oxidation state of 0, no longer +2. Iron (II) is used interior the call of a compound in which iron has an oxidation form of +2. again, it may't form FeF2 from the ions, on account that FeF2 is soluble in water. ============ extra follow up ============= ok, Erin. i will help you out of the morass of completely incorrect solutions that you've been getting. a million) H2(g) + I2(g) <==> 2HI(g) 2) Fe(s) + F2(g) --> FeF2(s)

Answer 4

first convert grams CO2 and grams H2O to moles

moles CO2 = wt/mw = 1.0562 g / 44.00 g/mole = 0.02400 moles
moles H2O = .5046 g / 18.02 g/mole = .02800 moles

next, calculate moles of C and moles of H

moles C = moles CO2 x moles C / mole CO2
= .02400 moles x ( 1 C / 1 CO2) = .02400 moles C
moles H = .02800 moles H2O x ( 2 H / 1 H2O) = .05600

so ratio of C to H is .02400 / .05600 is 3:7

and your empirical formula is C3H7

***update***
zor, 24:56 is not 3:8, it's 3:7 (24/8 = 3, 56/8 = 7)
sastro, nobody said this was an alkane

Answer 5

CnH(2n+2) + O2 ----> CO2 + H2O
From the weight of each component pleased comvert to molweight by diveded the valenci each molecule than you can calculate n=? pleased try it.

Answer 6

It is probely your homework I am not telling you that it is Co2