A 1.0542 gram sample of X was burned in a combustion apparatus. The results were
ID: 970305 • Letter: A
Question
A 1.0542 gram sample of X was burned in a combustion apparatus. The results were: Weight of ascribe tube before combustion 75.2382 g Weight of ascarite tube after combustion 76.8377 g Weight of drierite tube before combustion 81.4128 g Weight of drierite tube after combustion 81.7418 g Given this data, calculate the empirical formula of compound X: _____________ -------------------------------------------------------------------------------------------------------------------------------------------- A 1.4745 gram sample of Y was burned in a combustion apparatus. The results were: Weight of ascribe tube before combustion 80.7821 g Weight of ascarite tube after combustion 83.0196 g Weight of drierite tube before combustion 78.2988 g Weight of drierite tube after combustion 78.7560 g Given this data, calculate the empirical formula of compound Y: ______________
Explanation / Answer
A 1.0542 gram sample of X was burned in a combustion apparatus. The results were: Weight of ascribe tube before combustion 75.2382 g Weight of ascarite tube after combustion 76.8377 g Weight of drierite tube before combustion 81.4128 g Weight of drierite tube after combustion 81.7418 g Given this data, calculate the empirical formula of compound X: _____________
ascarite tubes always absorb carbon dioxide thus mass of CO2 produced in combustion = 76.8377 - 75.2382 = 1.5995g CO2
Now determine the moles of CO2
= 1.5995g/ mol / 44.0095 g/mol
0.0363 moles CO2
Every mol of CO2 has one mole of C thus 0.0363 moles CO2 have 0.0363 moles of C
the drierite tube absorbed water = 81.7418 - 81.4128 = 0.329g H2O
Now determine the moles of H2O:
0.329 g / 18.02 g/ mol
=0.01827 moles of H2O
Every mol of H2O has 2 mole of H thus 0.01827 moles H2O have 0.01827 X 2 moles of H = 0.03655 moles
Moles of C = 0.0363 moles
Moles of H = 0.03655 moles
molar ratioof C and H=
C : H =0 .0363 : 0.0365 or 1: 1
Thus the empirical formula is CH
A 1.4745 gram sample of Y was burned in a combustion apparatus. The results were: Weight of ascribe tube before combustion 80.7821 g Weight of ascarite tube after combustion 83.0196 g Weight of drierite tube before combustion 78.2988 g Weight of drierite tube after combustion 78.7560 g Given this data, calculate the empirical formula of compound Y: ______
ascarite tubes always absorb carbon dioxide thus mass of CO2 produced in combustion = 83.0196 – 80.7821= 2.2375 g CO2
Now determine the moles of CO2
= 2.2375 g/ mol / 44.0095 g/mol
0.051 moles CO2
Every mol of CO2 has one mole of C thus 0.051 moles CO2 have 0.051 moles of C
the drierite tube absorbed water = 78.7560-78.2988 = 0.4572g H2O
Now determine the moles of H2O:
0.4572 g / 18.02 g/ mol
=0.025 moles of H2O
Every mol of H2O has 2 mole of H thus 0.025 moles H2O have 0.025 X 2 moles of H = 0.050 moles
Moles of C = 0.051 moles
Moles of H = 0.050 moles
molar ratioof C and H=
C : H =0 .051: 0.050 or 1: 1
Thus the empirical formula is CH
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