**Sample Problem Analysis: Phosphorus Oxide**2. The analysis of a 1.00 g sample of phosphorus oxide, with a molar mass of about 284 g/mol, was found to contain 0.44 g of phosphorus and 0.56 g of oxygen. Calculate the empirical formula and the molecular formula for this phosphorus oxide. *(Show your calculations.)* ---**Explanation:**This problem involves determining both the empirical and molecular formulas of a compound based on its composition. The information provided includes the total mass of the sample, the mass of each element within the sample, and the molar mass of the compound.1. **Empirical Formula Calculation:** - **Mass of Phosphorus:** 0.44 g - **Mass of Oxygen:** 0.56 g Convert the mass of each element to moles: - Moles of Phosphorus = 0.44 g / (atomic mass of phosphorus, approximately 30.97 g/mol) - Moles of Oxygen = 0.56 g / (atomic mass of oxygen, approximately 16.00 g/mol) Determine the smallest whole number ratio of moles by dividing each by the smallest number of moles calculated.2. **Molecular Formula Calculation:** - Use the empirical formula mass to determine how many empirical units are in the molecular formula. - Divide the molar mass (284 g/mol) by the empirical formula mass to find the multiplication factor.This problem emphasizes stoichiometry and the relationship between atomic masses, molecular mass, and the composition of compounds. 2. The analysis of 1.00 g sample of phosphorus oxide, with a molar mass of about 284 g/mol, was found to contain 0.44 g of phosphorus and 0.56 g of oxygen. Calculate the empirical formula and the molecular formula for this phosphorus oxide. (Show your calculations).

Molar mass means the mass of one mole substance. Empirical formula is the smallest whole number…

2. The analysis of 1.00 g sample of phosphorus oxide, with a molar mass of about 284 g/mol, was found to contain 0.44 g of phosphorus and 0.56 g of oxygen. Calculate the empirical formula and the molecular formula for this phosphorus oxide. (Show your calculations).

2. The analysis of 1.00 g sample of phosphorus oxide, with a molar mass of about 284 g/mol, was found to contain 0.44 g of phosphorus and 0.56 g of oxygen. Calculate the empirical formula and the molecular formula for this phosphorus oxide. (Show your calculations).

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Q: A 45.00 g sample of tetraethyl lead contains 28.82 g lead, 13.37 g carbon and 2.81 g hydrogen.…

A: The given data: The mass of lead, (Pb)= 28.82 g The mass of carbon, (C)= 13.37 g The mass of…

Q: analysis of a sample of a compound shows that 0.131 g H, 4.61 g CI and 8.32 g of O are present…

Q: A compound of molar mass 177 g/mol contains only carbon, hydrogen, bromine, and oxygen. Analysis…

A: Assume hydrogen mass as x g and carbon mass as 8x g. The formula mass is calculated as follows, N1…

Q: Please provide all the work in detail because I will be using this to study

A: Number of moles of Mg and O can be calculated by the equation given below:

Q: Combustion analysis was performed on 1.50 g of a compound containing C, H and N to yield 3.21 g of…

A: Given, Mass of the compound = 1.50 g Mass of CO2 = 3.21 g Mass of H2O = 0.981 g Empirical…

Q: A sample of copper (II) nitrate hydrate has a mass of 1.278 g. The sample is heated until all water…

A: Given , mass of copper (II) nitrate hydrate= 1.278 g mass of anhydrous copper (II) nitrate =…

Q: A 0.858-g sample of copper was reacted with oxygen to form 1.074 g of copper oxide. Determine the…

A: Given, mass of copper sample = 0.858 g Mass of copper oxide formed = 1.074 g Empirical formula of…

Q: Complete the table. Element Moles Mass Cr 0.00442 Fe 73.5 mg Ti 1.009 x 10-3 Hg 1.78 kg

A: Mole is represented by the symbol mol. It is unit of measurement of substance. Mass is defined as…

Q: An unknown compound is processed using elemental analysis and found to contain 117.4 g of platinum,…

A: We know that, Number of mole = given mass/molar mass Atomic mass of platinum =195.084 g

Q: A 14.40 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis…

Q: A 12.95 g sample of Mo₂O3(s) is converted completely to another molybdenum oxide by adding oxygen.…

A: Given,Mass of Mo2O3 = 12.95 gMass of new molybdenum oxide after add oxygen = 13.82 g

Q: A 5.00g sample of a compound made of 4.42g mercury and bromine was found to contain 0.585g bromine.…

A: Given total mass of sample = 5.00 g mass of Mercury ( Hg ) = 4.42 g mass of bromine ( Br ) = 0.585…

Q: Calculate the empirical formula for 1.386 g Zn, 5.381 I

A: Given,

Q: A 12.3 g sample of a compound known to consist of only carbon and hydrogen was found to be 14.40 %…

A: Mass of sample = 12.3 g Hydrogen percent by mass = 14.40 % Mass % = Mass of solutemass of…

Q: The analysis of 1.00 g sample of phosphorous oxide, with a molar mass of about 284 g/mol, was found…

A: Given information: Mass of sample = 1.00 g Mass of phosphorus = 0.44 g Mass of oxygen = 0.56 g…

Q: A 2.86 g sample of a compound consisting of carbon, hydrogen, oxygen, nitrogen, and sulfur was…

A: Given, mass of 2.86 g sample of a compound produced 1.61 g CO2 and 0.989 g H2O.Moles of CO2 and H2O…

Q: on, hydrogen and oxygen 180.16 g/mol.

A: a) moles of C = ( 3.66g / 44.0gmol-1 ) = 0.83mol .......(1) Mass of C = ( moles of C ) • ( atomic…

Q: In a lab experiment, it was observed that 68.84 g of lead combines with 23.56 g of chlorine to form…

A: Given :- mass of lead = 68.84 g mass of chlorine = 23.56 g molar mass = 834 g To determine :-…

Q: Part D. A 28.50 g sample of a compound of carbon, sulfur, hydrogen, and combusted (burned). 35.25 g…

Q: Determine molecular formula. An organic compound that only contains C, H, and O was subjected to…

A: Since the compound is made up of only C, H and O. Hence we can assume the empirical formula of…

Q: An unknown sample is composed of hydrogen, sulfur, and oxgen. A 2.358 gram sample of this unknown…

A: Given data is: unknown sample (H,S;O) mass of sample = 2.358 g The reaction is: sample (H,S,O) + O2…

Q: A 11.21 g sample of Mo₂O3(s) is converted completely to another molybdenum oxide by adding oxygen.…

A: Number of mol = Mass/molar mass

Q: 1.651 g of Silver is combined with 0.1224 g of oxygen. Find the empirical formula.

A: moles of silver combining = mass / atomic mass atomic mass of Ag = 108 g/mol Hence moles of Silver =…

Q: Find the percentage composition of a compound that contains 1.94 g of carbon, 0.48 g of hydrogen,…

A: The percent composition of a compound is given by the ratio of the mass of each element to the total…

Q: Please provide all the work in detail because I will be using this to study

A: Consider the mass of compound is one gram. The percent of each element is converted to grams as,

Q: A 7.194 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis…

A: We have to predict the empirical and molecular formula.Given, The mass of organic compound = 7.194…

Q: A 27.129 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put…

A: Answer: Empirical formula is the formula containing all the atoms present in the compound in their…

Q: A 4.600 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis…

A: Empirical formula: An empirical formula is a chemical formula that does not indicate the entire…

Q: A 4.00 g sample of a hydrate of nickel(III) bromide looses 0.793 g of water when heated. Determine…

A: It is given that the mass of hydrate of nickel (III) bromide is 4 gOn heating it losses 0.793 g of…

Q: 7.00 g of iron is reacted with 3.00 g of oxygen. Find the empirical formula.

A: Since moles of Fe reaction = mass of Fe / atomic mass of Fe and moles of O reacting = mass of O /…

Q: 2. Chemical analysis of an ether shows 64.82% C, 13.60% H, and 21.58% O by mass. Determine the…

A: Given :- % C =64.82 % H =13.60 % O = 21.58 So, Grams of carbon = 64.82 g Grams of Hydrogen = 13.60 g…

Q: 3. Hydrated vanadium (II) sulfate, VSO4-XH2O, is 46.19% water by mass. Determine the empirical…

A: Given mass percent of water in hydrated salt = 46.19% Atomic mass of V = 50.94 amu Atomic mass of S…

Q: A chemical analysis determined that the Butyric acid molecule was made up of 54.5% C, 9.09% H, and…

A: Given: Percent mass of carbon = 54.5 % Percent mass of hydrogen = 9.09 % Percent mass of oxygen =…

Q: 2.00 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have…

A: Empirical formula :- The formula which gives the simplest whole number ratio of atoms of various…

Q: The rotten smell of a decaying animal carcass is partially due to a nitrogen-containing compound…

Q: When 2.666 grams of a hydrocarbon, C,H,, were burned in a combustion analysis apparatus, 7.803 grams…

A: Mass of hydrocarbon = 2.666 g Mass of CO2 = 7.803 g Mass of H2O = 4.792 g Molar mass of compund =…

Q: A 5.899 gram sample of an organic compound containing C, Hand O is analyzed by combustion analysis…

A: Mass of organic sample = 5.899 gMass of CO2 = 12.71 gMass of H2O = 5.204 gmolar mass of given…

Q: A sample of a white solid contains 0.212 g of magnesium and 0.140 g oxygen. What is the empirical…

Q: Fill in the name and empirical formula of each ionic compound that could be formed from the ions in…

A: An ionic compound is formed by the combination of a metal cation and a non-metal anion through…

Concept explainers

Mole Concept

Mole concept is a method to determine the amount of any substance that consists of some elemental particles using a grouping unit called ‘mole’. The mole concept enables one to handle the large amount of small elementary particles. The mole concept can be used in many calculations related to mole calculation, molar amount, mole ratio, and so on, which are significant in chemistry.

Question

**Sample Problem Analysis: Phosphorus Oxide**

2. The analysis of a 1.00 g sample of phosphorus oxide, with a molar mass of about 284 g/mol, was found to contain 0.44 g of phosphorus and 0.56 g of oxygen. Calculate the empirical formula and the molecular formula for this phosphorus oxide. *(Show your calculations.)*

---

**Explanation:**

This problem involves determining both the empirical and molecular formulas of a compound based on its composition. The information provided includes the total mass of the sample, the mass of each element within the sample, and the molar mass of the compound.

1. **Empirical Formula Calculation:**

- **Mass of Phosphorus:** 0.44 g
- **Mass of Oxygen:** 0.56 g

Convert the mass of each element to moles:
- Moles of Phosphorus = 0.44 g / (atomic mass of phosphorus, approximately 30.97 g/mol)
- Moles of Oxygen = 0.56 g / (atomic mass of oxygen, approximately 16.00 g/mol)

Determine the smallest whole number ratio of moles by dividing each by the smallest number of moles calculated.

2. **Molecular Formula Calculation:**

- Use the empirical formula mass to determine how many empirical units are in the molecular formula.
- Divide the molar mass (284 g/mol) by the empirical formula mass to find the multiplication factor.

This problem emphasizes stoichiometry and the relationship between atomic masses, molecular mass, and the composition of compounds.

2. The analysis of 1.00 g sample of phosphorus oxide, with a molar mass of about 284 g/mol, was found to contain 0.44 g of phosphorus and 0.56 g of oxygen. Calculate the empirical formula and the molecular formula for this phosphorus oxide. (Show your calculations).

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step 3

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 3 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions

Use percent composition and molar mass to determine molecular formula. Analysis of a compound of carbon, hydrogen and oxygen showed that it is 52.13% C and 13.15% H, with O accounting for the remainder. In a separate experiment, the molar mass of the compound was found to be 46.08 g/mol. Determine the molecular formula of the compound. Enter the elements in the order C, H, O.
Sample A 8.074 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 20.88 grams of CO₂ and 4.274 grams of H₂O are produced. In a separate experiment, the molar mass is found to be 136.2 g/mol. Determine the empirical formula and the molecular formula of the organic compound. (Enter the elements in the order C, H, O.) Empirical formula: Molecular formula:
When 4.007 grams of a hydrocarbon, C,Hy, were burned in a combustion analysis apparatus, 12.14 grams of CO, and 6.212 grams of H20 were produced. In a separate experiment, the molar mass of the compound was found to be 58.12 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon. Enter the elements in the order presented in the question. empirical formula = molecular formula =
When 6.608 grams of a hydrocarbon, C,Hy, were burned in a combustion analysis apparatus, 20.73 grams of CO2 and 8.489 grams of H,O were produced. In a separate experiment, the molar mass of the compound was found to be 70.13 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon. Enter the elements in the order presented in the question. empirical formula molecular formula
1) A fuel composed of carbon, hydrogen, and oxygen is fully combusted with 18.214 g of охудen to yield 23.118 g of carbon dioxide and 4.729 g of water. Find the empirical formula for the compound.
(please type answer no write by hend)
A 1.71 mg sample of copper reacts with oxygen to form 1.93 mg of a copper oxide compound, CuxOy. Determine the empirical formula of the copper oxide compound (CuxOy).