The compound known as butylated hydroxytoluene, abbreviated as bht, contains carbon, hydrogen, and oxygen. a 1.376 g sample of bht was combusted in an oxygen rich environment to produce 4.122 g of co 2 ( g ) and 1.350 g of h 2 o ( g ) . insert subscripts to complete the empirical formula of bht.

C15H24O

Explanation:

TO GET THE EMPIRICAL FORMULA, WE NEED TO KNOW THE MASSES AND CONSEQUENTLY THE NUMBER OF MOLES OF EACH OF THE INDIVIDUAL CONSTITUENT ELEMENTS.

FIRSTLY, WE CAN GET THE MASS OF THE CARBON FROM THAT OF THE CARBON IV OXIDE. WE NEED TO KNOW THE NUMBER OF MOLES OF CARBON IV OXIDE GIVEN OFF. THIS CAN BE CALCULATED BY DIVIVDING THE MASS BY THE MOLAR MASS OF CARBON IV OXIDE. THE MOLAR MASS OF CARBON IV OXIDE IS 44G/MOL

HENCE, THE NUMBER OF MOLES OF CARBON IV OXIDE IS 4.122/44 WHICH EQUALS 0.094. SINCE THERE IS ONLY ONE ATOM OF CARBON IN CO2 THEN THEY HAVE EQUAL NUMBER OF MOLES AND THUS THE NUMBER OF MOLES OF CARBON IS 0.094. WE CAN THEN PROCEED TO CALCULATE THE MASS OF CO2 PRESENT. THIS CAN BE CALCULATED BY MULTIPLYING THE NUMBER OF MOLES BY THE ATOMIC MASS UNIT. THE ATOMIC MASS UNIT OF CARBON IS 12. HENCE, THE MASS OF CO2 PRESENT IS 12 * 0.094 = 1.128gWE CAN NOW GET THE MASS OF THE HYDROGEN BY MULTIPLYING THE NUMBER OF MOLES OF WATER BY 2 AND ALSO ITS ATOMIC MASS UNITTO GET THE NUMBER OF MOLES OF WATER, WE SIMPLY DIVIDE THE MASS BY THE MOLAR MASS. THE MOLAR MASS OF WATER IS 18g/mol.  The NUMBER OF MOLES IS THUS 1.350/18 = 0.075THE NUMBER OF MOLES OF HYDROGEN IS TWICE THAT OF WATER SINCE IT CONTAINS 2 ATOMS PER MOLECULE OF WATER. ITS NUMBER OF MOLES IS THUS 0.075*2 = 0.15 MOLETHE MASS OF HYDROGEN IS THUS 0.075 * 2 * 1 = 0.15gWE CAN NOW FIND THE MASS OF OXYGEN BY SUBTRACTING THE MASSES OF HYDROGEN AND CARBON FROM THE TOTAL MASS.MASS OF OXYGEN = 1.376-0.15-1.128 = 0.098gTHE NUMBER OF MOLES OF OXYGEN IS THUS 0.098/16 = 0.006125WE CAN NOW USE THE NUMBER OF MOLES TO OBTAIN THE EMPIRICAL FORMULA.WE DO THIS BY DIVIDING EACH BY THE SMALLEST NUMBER OF MOLES WHICH IS THAT OF THE OXYGEN.C = 0.094/0.006125 = 15H = 0.15/0.006125 = 24O = 1THE EMPIRICAL FORMULA IS THUS C15H24O

It was ernest rutherford