Question

The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.944 g of H2O(g). Insert subscripts below to appropriately display the empirical formula of BHTa. n(CO2)=m/M=11.61/44.01=0.2632 molb. n(H2O)=m/M=3.0803/18.02=0.1709 molc. m(C)=m*M=0.2638*12.01=3.1682 gd. m(H)=m*M=0.1709 *2*1.008=0,3446ge. m(O)=3.879-3.1682-0.3446=0.3662 gf. n(O)= 0.3662/16.00=0.0227 molg. n(C):n(H):n(O)= 0.2638:2*0.1709:0.0227=12:15:1h. Empirical formula C12H15O

Answer 1

I believe here is the right question, so will just ignore the rest of the junk information from the previous message

The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.944 g of H2O(g). Insert subscripts below to appropriately display the empirical formula of BHT

Answer:

$C_{15}H_{24}0$

Explanation:

A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.994 g of H2O(g).

If all the carbon in BHT is present in $CO_2$ and also, all the hydrogen in BHT is  present in $H_2O$, Then we can determine for the corresponding numbers of moles of Carbon(C) and Hydrogen (H) respectively as:

moles of  $CO_2$ = $8.990 g*(\frac{1mole}{44.01g})$

                       =  0.2043 moles

∴ moles of C =  0.2043 moles

moles of $H_2O$ = $2.944 g *(\frac{1mole}{18.01g} )$

                       = 0.1635 moles

∴ moles of H = 2 × 0.1635 moles

                      = 0.327 moles

Since number of moles= $\frac{mass}{molarmass}$

number of moles of H =  0.327 moles

molar mass of H = 1.008 g/mol

∴  mass of H in the sample = 0.327 moles × 1.008 g/mol

                                             = 0.329616g

                                             

mass of C in the sample can be calculated as = 0.2043 moles × $(\frac{12.01g}{1 mole} )$

= 2.453643 g

mass of C+H in the sample = 2.453643g + 0.329616g

= 2.783259 g

mass of O can be calculated as = 3.001 g - 2.783259 g

= 0.217741 g

∴ moles of O = 0.217741g × $(\frac{1mole}{16.0g})$

= 0.0136 moles

Now, since; we've gotten our data, we can now proceed to calculate for the empirical formula.

C                                          H                                    O

0.2043                                0.327                             0.0136

Dividing by the least number (0.0136) , we have :

$\frac{0.2043}{0.0136}$                                     $\frac{0.327}{0.0136}$                               $\frac{0.0136}{0.0136}$

15.02                                      24.04                             1

≅

15                                             24                                  1

Therefore, the empirical formula would be : $C_{15}H_{24}0$