A solution containing a mixture of the compounds X and Y had an absorbance of 0.584 at 443 nm and an absorbance of 0.501 at 520 nm when measured with a 1.00 cm cell. The molar absorptivities (ε) of X and Y at each wavelength are shown in the table. What are the concentrations of X and Y in this mixture?| Wavelength (λ, nm) | Molar Absorptivity (ε, M⁻¹cm⁻¹) || ------------------ | ------------------------------ || | X | Y || 443 | 15790 | 3725 || 520 | 3841 | 6.180×10³ |In this setup, Beer's Law can be used to calculate the concentrations of compounds X and Y in the mixture. Beer's Law states: \[ A = ε \cdot c \cdot l \] where:- \( A \) is the absorbance,- \( ε \) is the molar absorptivity,- \( c \) is the concentration,- \( l \) is the path length of the cell (1.00 cm in this case).The information provided will help determine the concentrations of X and Y based on their unique absorbance at given wavelengths.

According to Beer Lambert's Law, A=ε.C.l where A is the absorption, ε is the molar absorptivity, C is the concentration and l is the path length. Absorption is additive in nature.Write the Beer lambert's expression for the mixture at 443 nm. A443=εxCxl +εyCyl Substitute 0.584 for A443, 15790 M-1.cm-1 for εx and 3725 M-1.cm-1 for εy and 1 cm for l and rewrite the expression. 0.584=(15790 M-1.cm-1)Cx(1 cm) +(3725 M-1.cm-1)Cy(1 cm) Simplify the equation. 0.584=(15790 M-1.Cx) +(3725 M-1.Cy) Substitute 0.501 for A520, 3841 M-1.cm-1 for εx and 6.18×103 M-1.cm-1 for εy and 1 cm for l and rewrite the expression. 0.501=(3841 M-1.Cx) +(6.18×103 M-1.Cy) Solve for Cx and Cy. Cx=2.09×10-5 MCy=6.81×10-5 M

Science

Chemistry

A solution containing a mixture of the compounds X and Y had an absorbance of 0.584 at 443 nm and an absorbance of 0.501 at 520 nm when measured with a 1.00 cm cell. The molar absorptivities (E) of X and Y at each wavelength are shown in the table. What are the concentrations of X and Y in this mixture? Wavelengt h (A, nm) Molar Absorptivity (E, M-'cm-1) Y 443 15790 3725 520 3841 6.180x103

A solution containing a mixture of the compounds X and Y had an absorbance of 0.584 at 443 nm and an absorbance of 0.501 at 520 nm when measured with a 1.00 cm cell. The molar absorptivities (E) of X and Y at each wavelength are shown in the table. What are the concentrations of X and Y in this mixture? Wavelengt h (A, nm) Molar Absorptivity (E, M-'cm-1) Y 443 15790 3725 520 3841 6.180x103

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 student prepared several solutions of an unknown molecule, each at a different concentration. The…

A: Given: Best fit line y= 0.00626x + 0.00732 Path length= 0.760 cm To find: molar absorptivity

Q: At 590 nm, crystal violet (a biological stain) has a molar absorptivity (ε) of 8.7×104 M-1cm-1.…

A: Spectrophotometry is a method to measure how much a substance absorbs light by measuring the…

Q: If triplicate absorbance readings of 0.419, 0.430, and 0.416 are recorded, what is the average %T…

A: The average absorbance can be calculated as follows-

Q: The molar absorptivity of a colored solution with a concentration of 0.0222M was determined to be…

Q: the absorbance of an equilibrium mixture of Fescn2+ was measured at 447nm and found to be 0.319.…

Q: The measured absorbance of a solution with a pathlength of 1.00 cm is 0.544. If the concentration is…

A: pathlength (l) =1.00 cm Absorbance (A) = 0.544 concentration(c) = 1.40 x 10 —3M

Q: Assuming that the sample optical path length is 1 cm, find the molar absorptivity at wavelength 1…

A: In this question a sample consists of hmr and Mr which show respective absorbance values. Together a…

Q: Maria was tasked to determine the concentration of a Fe2(SO4)3 solution. She prepared 5 different…

Q: Quinine in a 1.664 g antimalarial tablet was dissolved in sufficient 0.10 M HCl to give 500 mL of…

A: Given: Weight of the tablet, w = 1.664 g Molar concentration of HCl, M = 0.1 M Volume of the…

Q: solution with %T = 75.9% has an absorbance equal to: A = log (1 T) T= (% T 100)

A: Relationship between absorbance and transmittance, A = - log T And Relationship between absorbance…

Q: When measured in a 1.00 cm cuvette, a 4.tx10 M solution of species X exthibited absorbances of…

Concept explainers

UV Visible Spectroscopy

Spectroscopy refers to the absorption or emission of light or other radiations by matter. When a substance or chemical absorbs light or radiation, they reach an excited state followed by de-excitation or emission of the light or radiation forming distinct spectra. UV-visible spectroscopy or ultraviolet-visible refers to the absorption and emission in the ultraviolet and visible regions of the electromagnetic spectrum.

Question

A solution containing a mixture of the compounds X and Y had an absorbance of 0.584 at 443 nm and an absorbance of 0.501 at 520 nm when measured with a 1.00 cm cell. The molar absorptivities (ε) of X and Y at each wavelength are shown in the table. What are the concentrations of X and Y in this mixture?

| Wavelength (λ, nm) | Molar Absorptivity (ε, M⁻¹cm⁻¹) |
| ------------------ | ------------------------------ |
| | X | Y |
| 443 | 15790 | 3725 |
| 520 | 3841 | 6.180×10³ |

In this setup, Beer's Law can be used to calculate the concentrations of compounds X and Y in the mixture. Beer's Law states:
\[ A = ε \cdot c \cdot l \]
where:
- \( A \) is the absorbance,
- \( ε \) is the molar absorptivity,
- \( c \) is the concentration,
- \( l \) is the path length of the cell (1.00 cm in this case).

The information provided will help determine the concentrations of X and Y based on their unique absorbance at given wavelengths.

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

Trending now

This is a popular solution!

Step by step

Solved in 2 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

Calculate the molar absorptivity of a solute when its measured absorbance is 1.050 with a concentration of 1.45 x 10-3 M. Assume the path length of the cell is 1.0 cm.
What is the absorbance of a 2.33×10−4M2.33×10−4M solution of a compound with a molar absorptivity of 1.05×103M−1cm−11.05×103M−1cm−1 in a 1.00 cm cell?
6. Blue Blue dye stock solution 0.293 M Absorbance at 630 nm 0.00265 Calibration curve y = 0.0833x A solution is prepared by diluting 2.79 mL of the blue dye stock solution to 25.00 mL. The measured absorbance for the prepared solution is listed in the data table. (a) What is the theoretical molar concentration? [Blue]theoretical x 10 |M (b) What is the experimental molar concentration? [Blue]experimental x 10 M (c) What is the percent error? Percent error (blue) = %
The molar absorptivity of ascorbic acid is determined using a 0.50 mM (±0.04 mM) solution. The solution had an absorbance reading of 0.20±0.02 using a cell with a 1.00±0.01 cm. Calculate the uncertainty in the molar absorptivity of the ascorbic acid using propagation of errors and determine which measurement contributes the most to the error. Remember A = εbc
The standard curve was made by spectrophotographic analysis of equilibrated iron(III) thiocyanate solutions of known concentration. You are asked to analyze a Fe(SCN)2+ solution with an unknown concentration and an absorbance value of 0.392. The slope-intercept form of the equation of the line is y = 4538.1x +0.0077. The unknown was analyzed on the same instrument as the standard curve solutions at the same temperature. What is the Fe³+ concentration of the unknown solution? [Fe³+] = mol/L Absorbance Iron(III) thiocynate standard curve 1.0 V 0.5 0.0001 Fe³+ concentration (M) 0 0.0002
Blue Blue dye stock solution 0.293 M Absorbance at 630 nm 0.00265 Calibration curve y = 0.0833x A solution is prepared by diluting 2.81 mL of the blue dye stock solution to 25.00 mL. The measure absorbance for the prepared solution is listed in the data table. (a) What is the theoretical molar concentration? [Blue]theoretical | × 10 |M (b) What is the experimental molar concentration? [Bluelexperimental x 10 M (c) What is the percent error? Percent error (blue) =
The molar absorptivity constant of a particular chemical is 2.31 M/cm. Determine the concentration of a solution made from this chemical that has an absorbance of 0.408 with a cell path length of 1.12 cm.
A 5.00×10−4 M solution of analyte is placed in a sample cell that has a pathlength of 1.00 cm. At a wavelength of 490 nm, the solution’s absorbance is 0.338. What is the analyte’s molar absorptivity at this wavelength?
Suppose that a solution has an absorbance of 0.226 at a wavelength of 450 nm. If the concentration of the solution is 24 μM, what is the value of the molar absorptivity? The data were taken with a standard 1-cm cuvette.
The absorbance of an unknown dye solution is measured to be 1.06 by a spectrometer. The calibration plot of the spectrometer is provided below. What is the molar concentration of the unknown sample (in mole L)? Absorbance vs. Molarity 18 y3 67024x-0.0343 16 14 12 1 0.8 0.6 0.4 0.2 01 0.2 0.000005 00001 0000015 Concentration of solutions, mol/L 0.00002 0.000025 000003 0 163e-5 O 2.35e-6 0 163 0 235 Absorbance
Table 2: Absorbance Values of Standards and Unknowns. Sample Blank Standard Solution 1 Standard Solution 2 Standard Solution 3 Standard Solution 4 Standard Solution 5 Water Sample 1 Water Sample 2 Phosphate Concentration (in ppm) Y17 0 ppm 0.02 ppm 0.04 ppm 0.08 ppm 0.16 ppm 0.32 ppm Freeman Lake 0.001 0.325 0.292 0.413 0.315 0.039 0.054 0.049 Absorbance at 2= 690 nm Calculations: 1. Construct a phosphate standard curve in Excel by plotting concentration (in ppm) on your x-axis and Absorbance (unitless) on your y-axis for your known solutions. Label the axes on the graph and provide the curve with a title. Use a linear trendline to generate a best fit line to your data. Label the graph with the equation and the R" value. Insert your labeled graph in the space below. X
5. A solution of certain dye has a calibration graph shown below when measured at 425 nm in 1.00cm cells. A sample of the dye with an unknown concentration has an absorbance of 0.733. Determine the molar absorptivity of the dye at 425 nm and the concentration of the dye in the unknown solution. Absorbance vs. Concentration 1.200 1.100 1.000 0.900 0.800 0.700 0.600 0.500 0.400 1.50E-05 2.50E-05 3.50E-05 4.50E-05 5.50E-05 6.50E-05 Dye Concentration (Mol/L)