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REACH

Ethanol, 2,2'-oxybis-, reaction products with ammonia, morpholine derivs. residues

EC number: 272-712-1 | CAS number: 68909-77-3 The residuum from the reaction of diethylene glycol and ammonia. It consists predominantly of morpholine-based derivatives such as [(aminoethoxy)ethyl]morpholine, [(hydroxyethoxy)ethyl]morpholine, 3-morpholinone, and 4,4'-(oxydi-2,1-ethanediyl)bis[morpholine].

Life Cycle description
Uses advised against

Physical & Chemical properties

Vapour pressure

Currently viewing:

Administrative data

Link to relevant study record(s)

Reference

Endpoint:

vapour pressure

Type of information:

experimental study

Adequacy of study:

key study

Study period:

2010-10-04 - 2010-10-20

Reliability:

1 (reliable without restriction)

Rationale for reliability incl. deficiencies:

comparable to guideline study

Qualifier:

equivalent or similar to guideline

Guideline:

EU Method A.4 (Vapour Pressure)

GLP compliance:

yes (incl. QA statement)

Type of method:

effusion method: vapour pressure balance

Specific details on test material used for the study:

Sponsor's identification: Amine C8
Description: dark brown liquid
Chemical name: Ethanol, 2, 2-oxybis-, reaction products with ammonia, morpholine derivs. residues
Batch number: T7-271109
Date received: 21 September 2010
Expiry date: 31 December 2011
Storage conditions: room temperature in the dark

Key result

Temp.:

25 °C

Vapour pressure:

0.55 Pa

Transition / decomposition:

Run1

Temperature (ºC)	Temperature (K)	Reciprocal Temperature (K^-1)	Mass Difference (µg)	Mass Difference (kg)	Vapour Pressure (Pa)	Log₁₀Vp
21	294.15	0.003399626	248.31	2.483E-07	0.344717897	-0.462536169
22	295.15	0.003388108	279.46	2.795E-07	0.387962077	-0.411210724
23	296.15	0.003376667	312.71	3.127E-07	0.434121596	-0.362388609
24	297.15	0.003365304	350.35	3.504E-07	0.486375559	-0.313028257
25	298.15	0.003354016	390.59	3.906E-07	0.542238989	-0.265809259
26	299.15	0.003342805	437.31	4.373E-07	0.607098318	-0.216740970
27	300.15	0.003331667	488.13	4.881E-07	0.677649498	-0.168994879
28	301.15	0.003320604	543.65	5.437E-07	0.754725482	-0.122210986
29	302.15	0.003309614	605.45	6.055E-07	0.840519715	-0.075452095
30	303.15	0.003298697	675.64	6.756E-07	0.937961418	-0.027815025

A plot of Log₁₀(vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 1 gives the following statistical data using an unweighted least squares treatment.

Slope -4294.026
Standard deviation in slope 6.934

Intercept 14.137
Standard deviation in intercept 0.023

The results obtained indicate the following vapour pressure relationship:

Log₁₀(Vp (Pa)) = -4294.026/temp(K) + 14.137

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.265.

Run 1 - Graph of Log₁₀Vapour Pressure vs Reciprocal Temperature see Attachment 3 Run 1 to 6

Graph of Log₁₀Vapour Pressure vs Reciprocal Temperature.

Run 2

Temperature (ºC)	Temperature (K)	Reciprocal Temperature (K^-1)	Mass Difference (µg)	Mass Difference (kg)	Vapour Pressure (Pa)	Log₁₀Vp
20	293.15	0.003411223	237.63	2.376E-07	0.329891320	-0.481629111
21	294.15	0.003399626	261.59	2.616E-07	0.363153939	-0.439909241
22	295.15	0.003388108	291.54	2.915E-07	0.404732212	-0.392832229
23	296.15	0.003376667	322.69	3.227E-07	0.447976393	-0.348744872
24	297.15	0.003365304	357.24	3.572E-07	0.495940644	-0.304570298
25	298.15	0.003354016	398.18	3.982E-07	0.552775853	-0.257450936
26	299.15	0.003342805	443.20	4.432E-07	0.615275147	-0.210930627
27	300.15	0.003331667	494.42	4.944E-07	0.686381630	-0.163434348
28	301.15	0.003320604	547.14	5.471E-07	0.759570496	-0.119431913
29	302.15	0.003309614	609.74	6.097E-07	0.846475335	-0.072385692
30	303.15	0.003298697	679.33	6.793E-07	0.943084083	-0.025449585

A plot of Log₁₀(vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 2 gives the following statistical data using an unweighted least squares treatment.

Slope -4067.764
Standard deviation in slope 26.030

Intercept 13.389
Standard deviation in intercept 0.087

The results obtained indicate the following vapour pressure relationship:

Log₁₀(Vp (Pa)) = -4067.764/temp(K) + 13.389

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.255.

Run 2 - Graph of Log₁₀Vapour Pressure vs Reciprocal Temperature see Attachment 3 Run 1 to 6

Graph of Log₁₀Vapour Pressure vs Reciprocal Temperature.

Run 3

Temperature (ºC)	Temperature (K)	Reciprocal Temperature (K^-1)	Mass Difference (µg)	Mass Difference (kg)	Vapour Pressure (Pa)	Log₁₀Vp
20	293.15	0.003411223	236.43	2.364E-07	0.328225413	-0.483827797
21	294.15	0.003399626	261.09	2.611E-07	0.362459811	-0.440740140
22	295.15	0.003388108	290.74	2.907E-07	0.403621607	-0.394025593
23	296.15	0.003376667	321.80	3.218E-07	0.446740845	-0.349944339
24	297.15	0.003365304	358.64	3.586E-07	0.497884203	-0.302871653
25	298.15	0.003354016	397.08	3.971E-07	0.551248771	-0.258652366
26	299.15	0.003342805	442.51	4.425E-07	0.614317250	-0.211607289
27	300.15	0.003331667	492.13	4.921E-07	0.683202523	-0.165450539
28	301.15	0.003320604	546.64	5.466E-07	0.758876368	-0.119828971
29	302.15	0.003309614	605.95	6.060E-07	0.841213844	-0.075093589
30	303.15	0.003298697	678.44	6.784E-07	0.941848535	-0.026018933

A plot of Log₁₀(vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 3 gives the following statistical data using an unweighted least squares treatment.

Slope -4067.078
Standard deviation in slope 21.916

Intercept 13.385
Standard deviation in intercept 0.074

The results obtained indicate the following vapour pressure relationship:

Log₁₀(Vp (Pa)) = -4067.078/temp(K) + 13.385

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.256.

Run 3 - Graph of Log₁₀Vapour Pressure vs Reciprocal Temperature see Attachment 3 Run 1 to 6

Graph of Log₁₀Vapour Pressure vs Reciprocal Temperature.

Run 4

Temperature (ºC)	Temperature (K)	Reciprocal Temperature (K^-1)	Mass Difference (µg)	Mass Difference (kg)	Vapour Pressure (Pa)	Log₁₀Vp
20	293.15	0.003411223	236.23	2.362E-07	0.327947762	-0.484195329
21	294.15	0.003399626	261.09	2.611E-07	0.362459811	-0.440740140
22	295.15	0.003388108	289.55	2.896E-07	0.401969582	-0.395806809
23	296.15	0.003376667	321.30	3.213E-07	0.446046717	-0.350619653
24	297.15	0.003365304	357.04	3.570E-07	0.495662993	-0.304813505
25	298.15	0.003354016	396.38	3.964E-07	0.550276992	-0.259418645
26	299.15	0.003342805	440.21	4.402E-07	0.611124261	-0.213870475
27	300.15	0.003331667	489.63	4.896E-07	0.679731882	-0.167662359
28	301.15	0.003320604	543.45	5.435E-07	0.754447831	-0.122370786
29	302.15	0.003309614	605.45	6.055E-07	0.840519715	-0.075452095
30	303.15	0.003298697	674.74	6.747E-07	0.936711987	-0.028393922

A plot of Log₁₀(vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 4 gives the following statistical data using an unweighted least squares treatment.

Slope -4053.160
Standard deviation in slope 22.971

Intercept 13.337
Standard deviation in intercept 0.077

The results obtained indicate the following vapour pressure relationship:

Log₁₀(Vp (Pa)) = -4053.160/temp(K) + 13.337

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.257.

Run 4 - Graph of Log₁₀Vapour Pressure vs Reciprocal Temperature see Attachment 3 Run 1 to 6

Graph of Log₁₀Vapour Pressure vs Reciprocal Temperature.

Run 5

Temperature (ºC)	Temperature (K)	Reciprocal Temperature (K^-1)	Mass Difference (µg)	Mass Difference (kg)	Vapour Pressure (Pa)	Log₁₀Vp
20	293.15	0.003411223	233.93	2.339E-07	0.324754773	-0.488444458
21	294.15	0.003399626	261.09	2.611E-07	0.362459811	-0.440740140
22	295.15	0.003388108	287.95	2.880E-07	0.399748372	-0.398213296
23	296.15	0.003376667	320.10	3.201E-07	0.444380809	-0.352244705
24	297.15	0.003365304	355.94	3.559E-07	0.494135911	-0.306153583
25	298.15	0.003354016	395.68	3.957E-07	0.549305213	-0.260186280
26	299.15	0.003342805	439.31	4.393E-07	0.609874831	-0.214759290
27	300.15	0.003331667	489.33	4.893E-07	0.679315406	-0.167928536
28	301.15	0.003320604	542.35	5.424E-07	0.752920749	-0.123250734
29	302.15	0.003309614	608.35	6.084E-07	0.844545658	-0.073376866
30	303.15	0.003298697	671.35	6.714E-07	0.932005799	-0.030581385

A plot of Log₁₀(vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 5 gives the following statistical data using an unweighted least squares treatment.

Slope -4074.568
Standard deviation in slope 21.543

Intercept 13.408
Standard deviation in intercept 0.072

The results obtained indicate the following vapour pressure relationship:

Log₁₀(Vp (Pa)) = -4074.568/temp(K) + 13.408

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.258.

Run 5 -Graph of Log₁₀Vapour Pressure vs Reciprocal Temperature see Attachment 3 Run 1 to 6

Graph of Log₁₀Vapour Pressure vs Reciprocal Temperature.

Run 6

Temperature (ºC)	Temperature (K)	Reciprocal Temperature (K^-1)	Mass Difference (µg)	Mass Difference (kg)	Vapour Pressure (Pa)	Log₁₀Vp
20	293.15	0.003411223	234.13	2.341E-07	0.325032424	-0.488073314
21	294.15	0.003399626	258.00	2.580E-07	0.358170099	-0.445910673
22	295.15	0.003388108	280.46	2.805E-07	0.389350333	-0.409659449
23	296.15	0.003376667	318.90	3.189E-07	0.442714902	-0.353875859
24	297.15	0.003365304	353.15	3.532E-07	0.490262677	-0.309571168
25	298.15	0.003354016	393.78	3.938E-07	0.546667526	-0.262276724
26	299.15	0.003342805	438.01	4.380E-07	0.608070098	-0.216046353
27	300.15	0.003331667	487.34	4.873E-07	0.676552776	-0.169698320
28	301.15	0.003320604	541.45	5.415E-07	0.751671319	-0.123972021
29	302.15	0.003309614	604.05	6.041E-07	0.838576157	-0.076457490
30	303.15	0.003298697	669.65	6.697E-07	0.929645763	-0.031682506

A plot of Log₁₀(vapour pressure (Pa)) versus reciprocal temperature (1/T(K)) for Run 6 gives the following statistical data using an unweighted least squares treatment.

Slope -4102.766
Standard deviation in slope 35.676

Intercept 13.500
Standard deviation in intercept 0.120

The results obtained indicate the following vapour pressure relationship:

Log₁₀(Vp (Pa)) = -4102.766/temp(K) + 13.500

The above yields a vapour pressure (Pa) at 298.15 K with a common logarithm of -0.261.

Run 6 - Graph of Log₁₀Vapour Pressure vs Reciprocal Temperature see Attachment 3 Run 1 to 6

Graph of Log₁₀Vapour Pressure vs Reciprocal Temperature.

Conclusions:

The vapour pressure of the test item has been determined to be 0.55 Pa at at 25 ºC.

Executive summary:

INTRODUCTION

Hazardous physico-chemical properties of the test item have been determined. Methods employed were designed to be compatible with Commission Regulation (EC) No 440/2008 of 30 May 2008, Part A: Methods for the determination of physico-chemical properties

Method

The vapour pressure was determined using a vapour pressure balance with measurements being made at several temperatures and linear regression analysis used to calculate the vapour pressure at 25°C. Testing was conducted using a method designed to be compatible with Method A4 Vapour Pressure of Commission Regulation (EC) No 440/2008 of 30 May 2008.

Summary of Results

Run	Log₁₀[Vp(25ºC)]
1	-0.265
2	-0.255
3	-0.256
4	-0.257
5	-0.258
6	-0.261
Mean	-0.259
Vapour Pressure	5.5 x 10^-1Pa

The test item did not change in appearance under the conditions used in the determination.

Discussion

The unshaded point on run 1 has been omitted from the data analysis, because its position in relation to the other 10 points clearly indicates it is erroneous.

Conclusion

The vapour pressure of the test item has been determined to be 5.5 x 10^-1Pa at 25 ºC.

Description of key information

The vapour pressure of the test substance was determined to be 0.55 Pa at 25 °C using the effusion method (vapour pressure balance) following a GLP study compatible with EC test method A.4 (Atwal et al., 2010).

Key value for chemical safety assessment

Vapour pressure:: 0.55 Pa
at the temperature of:: 25 °C

Additional information

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