Question

Дано w(c)=87,5% Дco2=2,18 CxHy-?​

Answer 1

Решение:

w(C)=87.5%

w(H)=12.5%

w= (n*Ar(C))/Mr(CxHy)

Mr(CxHy) = 44*2.18=95.92

n(C) = w*Mr((CxHy) / 12

n(C)= 0.875*95.92/12 = 83.93/12 = 7

w(H)= 100-87.5=12.5 %

n(H) = 0.125*95.92 / 1 = 12

Ответ: C7H12

Answer 2

To determine the empirical formula of the hydrocarbon (CxHy) given the weight percent composition of carbon (w(c)) and the molar mass of carbon dioxide (Dco2), we can use the following steps:

Step 1: Calculate the moles of carbon (moles of C) in carbon dioxide (CO2): Molar mass of CO2 (Dco2) = 44.01 g/mol (carbon: 12.01 g/mol, oxygen: 16.00 g/mol) moles of C in CO2 = mass of C in CO2 / molar mass of C moles of C in CO2 = 2.18 g / 12.01 g/mol ≈ 0.1815 mol

Step 2: Calculate the moles of hydrogen (moles of H) in water (H2O): Molar mass of H2O = 18.015 g/mol (hydrogen: 1.01 g/mol, oxygen: 16.00 g/mol) moles of H in H2O = mass of H in H2O / molar mass of H moles of H in H2O = (87.5 - 2.18) g / 18.015 g/mol ≈ 4.7894 mol

Step 3: Find the ratio of moles of carbon to moles of hydrogen: The ratio of moles of C to moles of H can be found by dividing the moles of C by the moles of H: Ratio of C to H ≈ 0.1815 mol / 4.7894 mol ≈ 0.0379

Step 4: Convert the ratio to the simplest whole number ratio: To get the simplest whole number ratio, we multiply both moles by a factor that makes them close to whole numbers. In this case, we can multiply them by 1000 to get rid of the decimal places: Ratio of C to H ≈ 0.0379 * 1000 ≈ 37.9

Step 5: Round the ratio to the nearest whole number: Since we cannot have a fraction of an atom in the empirical formula, we round the ratio to the nearest whole number: Rounded ratio of C to H = 38

Step 6: Write the empirical formula using the rounded ratio: The empirical formula of the hydrocarbon is C38H38.

Please note that this empirical formula is a theoretical representation based on the given information. To get the actual molecular formula, you would need additional data, such as the molar mass or the molecular weight of the compound.