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a compound is found to be 30.45% n and 69.55 % o by mass. if 1.63 g of this compound occupy 389 ml at 0.00°c and 775 mm hg, what is the molecular formula of the compound?

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get a compound is found to be 30.45% n and 69.55 % o by mass. if 1.63 g of this compound occupy 389 ml at 0.00°c and 775 mm hg, what is the molecular formula of the compound? from EN Bilgi.

A compound is found to be 30.45 % N and 69.55 % O by mass. If 1.63 g of this compound occupy 389 mL at 0.00 degree C and 775 mmHg, what is the molecular formula of the compound?

Answer to: A compound is found to be 30.45 % N and 69.55 % O by mass. If 1.63 g of this compound occupy 389 mL at 0.00 degree C and 775 mmHg, what...

Question:

A compound is found to be 30.45 % N and 69.55 % O by mass. If 1.63 g of this compound occupy 389 mL at 0.00 C and 775 mmHg, what is the molecular formula of the compound?

Empirical and Molecular Formula:

The molecular formula of a compound represents the exact number of atoms of the elements making up the compound. When the numbers of atoms of the elements are written in a chemical formula in a way that they are expressed in the simplest whole-number ratio, then, the formula is referred to as the empirical formula of the compound.

Answer and Explanation: 1

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We assume that we have 100 g of the compound. Hence, from the given mass percents, the assumed mass contains 30.45 g N and 69.55 g O. Then, we convert...

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SOLVED:69) A compound is found to be 30.45% N and 69.55% 0 by mass If 1.63 g of this compound occupy 389 mL, at 0.00PC and 775 mm Hg; what is the molecular formula of the compound? A) NO2 C) N402 E) N204 B) Nzo D) N2O5

to determine the empirical formula of each of these compounds from the percent mass. We simply need to assume we have 100 g of the compound. Well, we have 100 g of the compound, then at 71 89% Valium we have 71 89 g of Valium which we can divide by the molar mass of Valium to get molds of Valium in the compound .352 We have 28.11% bro. mean 100g of the compound has 28 .11g of Romaine, Dividing that by the Molar Mass, 79.904 gives us .352 moles of roaming in the compound. So with .352 moles of Romaine, a thallium in .352 moles of bromine. In the compound, we have equal moles of both of them, dividing both by one, we get one mould thallium for every one mole of romaine. So the empirical formula is T lbr and the name is thallium bromide. We may want to say thallium one bromide as the charges one and thallium is a metal, it's not a transition metal. So commonly the roman numeral is not used but neither is lead and a roman numeral is used with lead. So it's correct to say Valium one bromide. For the next compound, it's 74 5, 1% lead. So if we have 100 g of the compound, we have 74.51 g of lead, which we can divide by the molar mass have led to get the moles of lead in the compound .3596. If that same compound is 25.49% chlorine, then 100 g of the compound has 25.49 g of chlorine, which we can divide by the molar mass in order to get the moles of chlorine in the compound. So we have 3596 moles of lead. For every 7190 moles of chlorine, dividing by the smaller of the two. We get one mole of lead for every two moles of chlorine. So the empirical formula is PVC L two and we would name it led to chloride chlorine has one minus charge, chloride does. There are two of them. So let it has to have a two plus charge. So we name it led to chloride for the next compound, that's 82.24% nitrogen, 100 g of the compound is 82.24 g of nitrogen, divide that by the molar mass. To get the moles of nitrogen in the compound and if that same compound is 17.76% hydrogen, then 100 g of the compound has 17.7 g of hydrogen, which we can convert into moles of hydrogen by divided by the molar mass of hydrogen And we get 5.871 moles of nitrogen for every 17 .62 moles of hydrogen dividing each of these by the smaller of the two. In an attempt to give us whole numbers, gives us one more nitrogen for every three moles of hydrogen. So the empirical formula is N. H three and the common name is ammonia. You might also call it nitrogen hydride, but hardly anybody ever does. The last compound is 72.24% magnesium. So 100g of the compound has 72.24g of magnesium, Which when divided by the molar mass of Magnesium, gives us the moles of Magnesium in the compound 2.972. That same compound has 27.76 moles of nitrogen. So 100 g has 70 27.76 g of nitrogen, which were divided by the molar mass of nitrogen gives us the moles of nitrogen in the compound 1982 moles of nitrogen for every 2.972 moles of magnesium divided by the smaller of the two. And in an attempt to get whole numbers, gives us one more nitrogen for everyone. Five moles of magnesium. We can then multiply both of these by two, which will give us whole numbers three moles magnesium for every two moles, nitrogen. Thus the empirical formula is MG three and 2 and the name of the compound is magnesium nitride

Melissa M.

Chemistry 101

3 months, 2 weeks ago

69) A compound is found to be 30.45% N and 69.55% 0 by mass If 1.63 g of this compound occupy 389 mL, at 0.00PC and 775 mm Hg; what is the molecular formula of the compound? A) NO2 C) N402 E) N204 B) Nzo D) N2O5

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Given the following mass percentages of the elements in certain compounds, determine the empirical formula in each case and name the compound. (a) 71.89% T128.11%Br (b) 74.51% Pb25.49%Cl (c) 82.24% N17.76%H (d) 72.24%Mg27.76% N

General Chemistry 4th

Chapter 11

Chemical Calculations

Video Transcript

to determine the empirical formula of each of these compounds from the percent mass. We simply need to assume we have 100 g of the compound. Well, we have 100 g of the compound, then at 71 89% Valium we have 71 89 g of Valium which we can divide by the molar mass of Valium to get molds of Valium in the compound .352 We have 28.11% bro. mean 100g of the compound has 28 .11g of Romaine, Dividing that by the Molar Mass, 79.904 gives us .352 moles of roaming in the compound. So with .352 moles of Romaine, a thallium in .352 moles of bromine. In the compound, we have equal moles of both of them, dividing both by one, we get one mould thallium for every one mole of romaine. So the empirical formula is T lbr and the name is thallium bromide. We may want to say thallium one bromide as the charges one and thallium is a metal, it's not a transition metal. So commonly the roman numeral is not used but neither is lead and a roman numeral is used with lead. So it's correct to say Valium one bromide. For the next compound, it's 74 5, 1% lead. So if we have 100 g of the compound, we have 74.51 g of lead, which we can divide by the molar mass have led to get the moles of lead in the compound .3596. If that same compound is 25.49% chlorine, then 100 g of the compound has 25.49 g of chlorine, which we can divide by the molar mass in order to get the moles of chlorine in the compound. So we have 3596 moles of lead. For every 7190 moles of chlorine, dividing by the smaller of the two. We get one mole of lead for every two moles of chlorine. So the empirical formula is PVC L two and we would name it led to chloride chlorine has one minus charge, chloride does. There are two of them. So let it has to have a two plus charge. So we name it led to chloride for the next compound, that's 82.24% nitrogen, 100 g of the compound is 82.24 g of nitrogen, divide that by the molar mass. To get the moles of nitrogen in the compound and if that same compound is 17.76% hydrogen, then 100 g of the compound has 17.7 g of hydrogen, which we can convert into moles of hydrogen by divided by the molar mass of hydrogen And we get 5.871 moles of nitrogen for every 17 .62 moles of hydrogen dividing each of these by the smaller of the two. In an attempt to give us whole numbers, gives us one more nitrogen for every three moles of hydrogen. So the empirical formula is N. H three and the common name is ammonia. You might also call it nitrogen hydride, but hardly anybody ever does. The last compound is 72.24% magnesium. So 100g of the compound has 72.24g of magnesium, Which when divided by the molar mass of Magnesium, gives us the moles of Magnesium in the compound 2.972. That same compound has 27.76 moles of nitrogen. So 100 g has 70 27.76 g of nitrogen, which were divided by the molar mass of nitrogen gives us the moles of nitrogen in the compound 1982 moles of nitrogen for every 2.972 moles of magnesium divided by the smaller of the two. And in an attempt to get whole numbers, gives us one more nitrogen for everyone. Five moles of magnesium. We can then multiply both of these by two, which will give us whole numbers three moles magnesium for every two moles, nitrogen. Thus the empirical formula is MG three and 2 and the name of the compound is magnesium nitride

What is the molecular formula for the compound that is made up of 30.45 g N and 69.55 g O, and has a formula mass of 92.02 amu?

"N"_2"O"_4 Moles of nitrogen "n"_"N" = "30.45 g"/"14 g/mol" = "2.175 mol" Moles of oxygen "n"_"O" = "69.55 g"/"16 g/mol" = "4.347 mol" Ratio of moles of nitrogen and oxygen "n"_"N"/"n"_"O" = "2.175 mol"/"4.347 mol" ≈ 1/2 ∴ Emperical formula "= NO"_2 "Molecular formula mass"/"Emperical formula mass" = "92.02 amu"/"46.0 amu" ≈ 2 ∴ Molecular formula = 2 × "NO"_2 = color(Green)("N"_2"O"_4)

Moles of nitrogen

nN=30.45 g14 g/mol=2.175 mol

Moles of oxygen

nO=69.55 g16 g/mol=4.347 mol

Ratio of moles of nitrogen and oxygen

nNnO=2.175 mol4.347 mol≈12

∴ Emperical formula = NO2

Molecular formula massEmperical formula mass=92.02 amu46.0 amu≈2

∴ Molecular formula =2×NO2=N2O4

Source : socratic.org

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James 6 month ago

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