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noname [10]
2 years ago
11

A sample of a compound is determined to have 1.17 g of carbon and 0.287 g of hydrogen. what is the correct representation of the

empirical formula for the compound?
Chemistry
1 answer:
yarga [219]2 years ago
5 0

CH3 is the empirical formula for the compound.

A sample of a compound is determined to have 1.17g of Carbon and 0.287 g of hydrogen.

The number of atom or moles in the compound is

1.17 g C X  1 mol of C / 12.011 g C = 0.097411 mol of C.

0.287 g H x 1 mol of  H / 1 g H = 0.28474 mol H.

This compound contains 0.097411 mol of carbon and 0.28474 mol of Hydrogen.

So we can represent the compound with the formula C0.974H0.284.

Subscripts in formulas can be made into whole numbers by multiplying the smaller subscript by the larger subscript.

we can divide 0.284 by 0.0974.

0.284 / 0.0974 = 3.

So here, Carbon is one and hydrogen is 3.

We can write the above formula as a CH3.

Hence the empirical formula for the sample compound is CH3.

For a detailed study of the empirical formula refer given link brainly.com/question/13058832.

#SPJ1.

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Answer:

Concentration of Cr_2O_7^{2-} = 0.03101 M

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Explanation:

A)

The reduction for Cr_2O_7^{2-} is;

Cr_2O_7^{2-} + 14 H ^+ _{(aq)}  + 6 e^- -----> 2 Cr^{3+} _{(aq)}+7H_2O _{(l)}

Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-

6 moles of Cu ^+ = 1 mole of Cr_2O_7^{2-}

number of moles of Cu reacted = \frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }

number of moles of Cu reacted = \frac{0.4749}{63.55}

number of moles of Cu reacted = 0.00747 mole

number of moles of Cr_2O_7^{2-}reacted = \frac{0.00747}{6}

number of moles of Cr_2O_7^{2-}reacted = 0.001245 mole

Concentration of Cr_2O_7^{2-} = \frac{number \ of moles }{Volume}

Given that the volume = 40.15 mL = 40.15 *10^{-3}; we have:

Concentration of Cr_2O_7^{2-} = \frac{0.001245}{40.15*10^{-3}}

Concentration of Cr_2O_7^{2-} = 0.03101 M

B)

The reduction for MnO_4^- is;

MnO_4^- + 8H^+ + 5 e^- -----> Mn^{2+} + 4H_2O

Cu^+_{(aq)} -----> Cu^{2+} _{(aq)} + 1 e^-

5 moles of Cu ^+ = 1 mole of Cr_2O_7^{2-}

number of moles of Cu reacted = \frac{mass \ of \ Cu \ wire }{ molecular weigh tof \ Cu wire }

number of moles of Cu reacted = \frac{0.4749}{63.55}

number of moles of Cu reacted = 0.00747 mole

number of moles of MnO_4^- reacted = \frac{0.00747}{5}

number of moles of MnO_4^- reacted = 0.001494 mole

Concentration of MnO_4^- = \frac{number \ of moles }{Volume}

Given that the volume = 40.15 mL = 40.15 *10^{-3}; we have:

Concentration of MnO_4^- = \frac{0.001494 }{40.15*10^{-3}}

Concentration of MnO_4^- = 0.03721 M

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7 0
2 years ago
In the preparation of a certain alkyl halide, 10 g of sodium bromide (NaBr), 10 mL distilled water (H20), and 9 mL 3-methyl-1-bu
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Percentage yield shows the amount of reactants converted into products. The percentage yield of the reaction is 51.7%.

The equation of the reaction is sown in the image attached. The reaction is 1:1 as we can see.

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We can obtain the mass of 3-methyl-1-butanol from its density.

Mass = density × volume

Density of 3-methyl-1-butanol =  0.810 g/mL

Volume of  3-methyl-1-butanol = 9 mL

Mass of 3-methyl-1-butanol = 0.810 g/mL × 9 mL

Mass of 3-methyl-1-butanol = 7.29 g

Number of moles of 3-methyl-1-butanol =  mass/molar mass =  7.29 g/88 g/mol = 0.083 moles

Since the reaction is 1:1 then the limiting reagent is 3-methyl-1-butanol

Mass of product 1-bromo-3-methylbutane = number of moles × molar mass

Molar mass of 1-bromo-3-methylbutane = 151 g/mol

Mass of product 1-bromo-3-methylbutane = 0.083 moles × 151 g/mol

= 12.53 g

Recall that % yield = actual yield/theoretical yield × 100

Actual yield of product = 6.48 g

Theoretical yield = 12.53 g

% yield = 6.48 g/12.53 g × 100

% yield = 51.7%

Learn more: brainly.com/question/5325004

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3 years ago
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