Question

M(CxHy)=5,8г m(CO2)=17,6г m(H20)=9г Д(N2)=2,07 формула-?

Answer 1

По D находим молярную массу соединения. Затем для проверки просто можно подставить один из значений.

Answer 2

To determine the chemical formula of the compound, we need to first find the moles of each element present in the compound.

We know that the mass of carbon in the compound is given by: m(C) = m(CxHy) = 5.8 g

Using the molar mass of carbon (12.01 g/mol), we can calculate the moles of carbon: n(C) = m(C) / M(C) = 5.8 g / 12.01 g/mol = 0.483 moles

Similarly, the mass of hydrogen is given by: m(H) = m(CxHy) x y = 5.8 g x y

Using the molar mass of hydrogen (1.01 g/mol), we can calculate the moles of hydrogen: n(H) = m(H) / M(H) = 5.8 g x y / 1.01 g/mol = 5.74y moles

The mass of oxygen in the compound can be calculated from the masses of carbon dioxide and water produced: m(O) = m(CO2) + m(H2O) - 2m(N2) = 17.6 g + 9 g - 2(2.07 g) = 22.46 g

Using the molar mass of oxygen (16.00 g/mol), we can calculate the moles of oxygen: n(O) = m(O) / M(O) = 22.46 g / 16.00 g/mol = 1.404 moles

Now, we need to find the ratio of moles of each element to get the empirical formula. Dividing the moles of each element by the smallest number of moles (0.483) gives:

n(C) = 0.483 / 0.483 = 1 n(H) = 5.74y / 0.483 n(O) = 1.404 / 0.483

To simplify the calculation, we can assume that y is a small integer value (typically 1, 2, or 3) that makes n(H) a whole number. By inspection, y=4 gives n(H) = 11, which is a whole number. Therefore:

n(C) = 1 n(H) = 11 n(O) = 29

The empirical formula is then CH11O29. To get the molecular formula, we need to determine the molecular mass of the compound. We know that:

m(CxHyOz) = m(CxHy) + m(O) = 5.8 g + 22.46 g = 28.26 g

Using the molar mass of the empirical formula (CH11O29), we can calculate the factor by which the empirical formula needs to be multiplied to obtain the molecular formula:

factor = molar mass of molecular formula / molar mass of empirical formula = (28.26 g/mol) / (12.01 g/mol + 11.01 g/mol + 29.00 g/mol) = 2.00 (approx.)

Multiplying the subscripts of the empirical formula by the factor gives the molecular formula:

C2H22O58

Therefore, the chemical formula of the compound is C2H22O58.