Chemistry, 03.07.2019 03:40 owlette2001

Combustion analysis of 1.200 g of an unknown compound containing carbon, hydrogen, and oxygen produced 2.086 g of co2 and 1.134 g of h2o. what is the empirical formula of the compound?

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the empirical formula is C3H8O2 You need to determine the relative number of moles of hydrogen and carbon. So you first calculate the molar mass of CO2 and H20 Atomic weight of carbon = 12.0107 Atomic weight of hydrogen = 1.00794 Atomic weight of oxygen = 15.999 Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 Molar mass H2O = 2 * 1.00794 + 15.999 = 18.01488 Now calculate the number of moles of CO2 and H2O you have Moles CO2 = 2.086 g / 44.0087 g/mole = 0.0474 mole Moles H2O = 1.134 g / 18.01488 g/mole = 0.062948 mole Calculate the number of moles of carbon and hydrogen you have. Since there's 1 carbon atom per CO2 molecule, the number of moles of carbon is the same as the number of moles of CO2. But since there's 2 hydrogen atoms per molecule of H2O, The number of moles of hydrogen is double the number of moles of H2O Moles Carbon = 0.0474 Moles Hydrogen = 0.062948 * 2 = 0.125896 Now we need to determine how much oxygen is in the compound. Just take the mass of the compound and subtract the mass of carbon and hydrogen. What's left will be the mass of oxygen. Then divide that mass by the atomic weight of oxygen to get the number of moles of oxygen we have. 1.200 - 0.0474 * 12.0107 - 0.125896 * 1.00794 = 0.503797 Moles oxygen = 0.503797 / 15.999 = 0.031489 So now we have a ratio of carbon:hydrogen:oxygen of 0.0474 : 0.125896 : 0.031489 We need to find a ratio of small integers that's close to that ratio. Start by dividing everything by 0.031489 (selected because it's the smallest value) getting 1.505288 : 3.998095 : 1 The 1 for oxygen and the 3.998095 for hydrogen look close enough. But the 1.505288 for carbon doesn't work. But it looks like if we double all the numbers, we'll get something close to an integer for everything. So do so. 3.010575 : 7.996189 : 2 Now this looks good. Rounding everything to an integer gives us 3 : 8 : 2 So the empirical formula is C3H8O2

Answer from: Quest

tracks, burrows, eggshells, nests, tooth marks, or gastroliths

Answer from: Quest

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

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